Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

No Downloads

Total views

1,264

On SlideShare

0

From Embeds

0

Number of Embeds

1

Shares

0

Downloads

76

Comments

0

Likes

2

No embeds

No notes for slide

- 1. ENGINEERING CURVES By: Prof K. M.Joshi Assi. Professor, MED, SSAS Institute of Technology, Surat.
- 2. ENGINEERING CURVESELLIPSE HYPERBOLA PARABOLA1.Concentric Circle 1.Rectangular 1.Rectangle Method Method Hyperbola (coordinates given) 2 Method of Tangents2.Rectangle Method (Triangle Method) 2 Rectangular3.Oblong Method Hyperbola 3.Basic Locus Method (P-V diagram - (Directrix – focus)4.Arcs of Circle Method Equation given) www.joshikandarp.webs.com5.Rhombus Metho 3.Basic Locus Method (Directrix – focus)6.Basic Locus Method (Directrix – focus)
- 3. CONIC SECTIONSELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONSBECAUSE THESE CURVES APPEAR ON THE SURFACE OF A CONEWHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES. Ellipse Section Plane Section Plane Hyperbola ola Parallel to Axis.Through Generators rab Pa www.joshikandarp.webs.com Section Plane Parallel to end generator.
- 4. : Common Definition of Ellipse, Parabola & HyperbolaThese are the loci of points moving in a plane such that theratio of it’s distances from a fixed point And a fixed linealways remains constant. The Ratio is called ECCENTRICITY. (E) For Ellipse E<1 For Parabola E=1 For Hyperbola E>1 It is a locus of a point moving in a plane such that the SUMof it’s distances from TWO fixed points always remains www.joshikandarp.webs.comconstant. [And this sum equals to the length of major axis. TheseTWO fixed points are FOCUS 1 & FOCUS 2]
- 5. Problem 1 : Draw ellipse by concentric circle method. Take major axis 100 mm and minor axis 70 mm long.Steps: 3 2 41.Draw both axes as perpendicularbisectors of each other & name theirends as shown. C2. Taking their intersecting point as 1 5a center, draw two concentric circles 3 2 4considering both as respectivediameters. 1 53. Divide both circles in 12 equalparts & name as shown. A B4. From all points of outer circledraw vertical lines downwards andupwards respectively. 10 65. From all points of inner circle 10 9draw horizontal lines to intersect 7 6 www.joshikandarp.webs.com 8those vertical lines. D6. Mark all intersecting pointsproperly as those are the points onellipse. 97. Join all these points along with 8 7the ends of both axes in smoothpossible curve. It is required ellipse. ELLIPSE CONCENTRIC CIRCLE METHOD
- 6. Steps:1 Draw a rectangle takingmajor and minor axes as sides. Problem 2 : Draw ellipse by Rectangle2. In this rectangle draw both method. Take major axis 110 mm andaxes as perpendicular bisectors minor axis 75 mm long.of each other..3. For construction, selectupper left part of rectangle.Divide vertical small side andhorizontal long side into same Dnumber of equal parts. (here 4 4divided in four parts) 3 34. Name those as shown.5. Now join all vertical points 2 21,2,3,4, to the upper end ofminor axis. And all horizontal 1 1points i.e.1,2,3,4 to the lowerend of minor axis. A B 1 2 3 4 3 2 16. Then extend C-1 line up toD-1. and mark that point.Similarly extend C-2, C-3, C-4lines up to D-2, D-3, & D-4 www.joshikandarp.webs.comlines.7. Mark all these pointsproperly and join all along withends A and D in smooth Cpossible curve. Do similarconstruction in right side partalong with lower half of therectangle. Join all points in ELLIPSEsmooth curve. It is required BY RECTANGLE METHODellipse.
- 7. Problem 3: Draw ellipse by Oblong method. Draw aparallelogram of 100 mm and 70 mm long sides with includedangle of 50. Inscribe Ellipse in it.Steps Are Similar To The Previous Case (Rectangle Method) Only InPlace Of Rectangle, Here Is A Parallelogram. D 4 4 3 3 2 2 1 1 A 1 2 3 4 3 2 1 B www.joshikandarp.webs.com C ELLIPSE BY OBLONG METHOD
- 8. PROBLEM 4: MAJOR AXIS AB & MINOR AXIS CD ARE 100 AMD 70MM LONG RESPECTIVELY. DRAW ELLIPSE BY ARCS OF CIRLES METHOD.STEPS: As per the definition Ellipse is locus of point P moving in a plane such that the SUM of it’s distances from two fixed1. Draw both axes as usual. points (F1 & F2) remains constant and equals to the length ofName the ends & intersecting major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB)point2. Taking AO distance i.e. halfmajor axis, from C, mark F1 & p4 CF2 On AB. ( focus 1 and 2.) p33. On line F1- O taking any p2distance, mark points 1,2,3, & 4 p14. Taking F1 center, withdistance A-1 draw an arc aboveAB and taking F2 center, with B-1 distance cut this arc. Name A B Othe point p1 F1 1 2 3 4 F25. Repeat this step with same www.joshikandarp.webs.comcenters but taking now A-2 & B-2 distances for drawing arcs.Name the point p26. Similarly get all other Ppoints. With same stepspositions of P can be located Dbelow AB.7. Join all points by smooth ELLIPSEcurve to get an ellipse/ BY ARCS OF CIRCLE METHOD
- 9. PROBLEM 5: DRAW RHOMBUS OF 100 MM & 70 MM LONG DIAGONALS AND INSCRIBE AN ELLIPSE IN IT. 2STEPS:1. Draw rhombus of given dimensions. A B2. Mark mid points of all sides& name Those A,B,C,& D 3 43. Join these points to the endsof smaller diagonals.4. Mark points 1,2,3,4 as four D C centers.5. Taking 1 as center and 1-A www.joshikandarp.webs.com radius draw an arc AB. 16. Take 2 as center draw anarc CD.7. Similarly taking 3 & 4 as ELLIPSEcenters and 3-D radius draw BY RHOMBUS METHODarcs DA & BC.
- 10. PROBLEM 6: A ball thrown in air attains 100 m height and covershorizontal distance 150 m on ground. Draw the path of the ball(projectile) 6 6 5 5 4 4 3 3 2 2 www.joshikandarp.webs.com 1 1 1 2 3 4 5 6 5 4 3 2 1 PARABOLA RECTANGLE METHOD
- 11. Problem 7: Draw an isosceles triangle of 100 mm long base and 110mm long altitude. Inscribe a parabola in it by method of tangents.Steps: C 14 11. Construct triangle as per the givendimensions. 13 2 122. Divide it’s both sides in to same no. of 3 11equal parts. 4 10 53. Name the parts in ascending and 9descending manner, as shown. 6 8 74. Join 1-1, 2-2,3-3 and so on. 7 8 6 95. Draw the curve as shown i.e. tangent toall these lines. The above all lines being 5 10 tangents to the curve, it is called method 4 11 www.joshikandarp.webs.comof tangents. 3 12 2 13 1 14 A B PARABOLA METHOD OF TANGENTS
- 12. STEPS : Ellipse / Parabola / Hyperbola DIRECTRX FOCUS METHOD1.Locate the directrix line and focal point bygiven distance (say 50 mm). Divide thisdistance in such a way that vertex point V Adivide this distance in given ratio i.e. 5’eccentricity, (say e = 2/3 = VF / VC ) 4’ 3’2.Draw the line from V parallel to directrix. DIRECTRIX 2’Take V as center and VF radius to draw arc 1’intersecting with this line. P2 P13.Draw the line passing from this point ofintersection and point C. (this line is tangent tothe curve exactly above point F)4.Now dived the VF and beyond in equal parts (vertex) V F (focus)i.e. 1, 2, 3…. etc and draw the parallel line C 1 2 3 4 5from each point and also note 1’, 2’, 3’… asshown in figure.5.Take distance 1-1’ , F as center and tick markthat on both side of line passing from 1; get P1’point P1 and P1’ Repeat same, i.e. distance 2-2’, P2’F center and tick mark on line passing from 2;get point P2 and P2’.6.Similarly repeat this process to get points P3 BP4.. Etc.7.6.Join all these points in smooth curve.
- 13. PROBLEM 8:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN APLANESUCH THAT THE RATIO OF IT’S DISTANCES FROM F AND LINE AB REMAINSCONSTANTAND EQUALS TO 2/3 DRAW LOCUS OF POINT P. { ECCENTRICITY = 2/3 } ELLIPSE STEPS: A 1. Draw a vertical line AB and point F DIRECTRIX 50 mm from it. 45mm 2. Divide 50 mm distance in 5 parts. 30mm 3. Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. It is first point giving ratio of it’s distances from F and AB 2/3 i.e 20/30 (vertex)V F ( focus) 4. Form more points giving same ratio such as 30/45, 40/60, 50/75 etc. 5. Taking 45,60 and 75mm distances from line AB, draw three vertical lines www.joshikandarp.webs.com to the right side of it. 6. Now with 30, 40 and 50mm distances in compass cut these lines above and below, with F as center. B 7. Join these points through V in smooth curve. This is required locus of P. It is an ELLIPSE. ELLIPSE DIRECTRIX-FOCUS METHOD
- 14. Steps: 21) Extend horizontal line fromP to right side. HYPERBOLA2) Extend vertical line from P THROUGH A POINTupward. OF KNOWN CO-ORDINATES3) On horizontal line from P,mark some points taking anydistance and name them afterP-1, 2,3,4 etc. 14) Join 1-2-3-4 points to pole O.Let them cut part [P-B] also at1,2,3,4 points.5) From horizontal 1,2,3,4 draw 2 1 P 1 2 3vertical lines downwards and6) From vertical 1,2,3,4 points[from P-B] draw horizontal 1lines. www.joshikandarp.webs.com7) Line from 1 horizontal and 40 mm 2line from 1 vertical will meet atP1.Similarly mark P2, P3, P4 3points.8) Repeat the procedure by Omarking four points on upwardvertical line from P and joining 30 mmall those to pole O. Name thispoints P6, P7, P8 etc. and jointhem by smooth curve.
- 15. STEPS: A1. Draw a vertical line AB and point F 50 mm from it. 30mm2.Divide 50 mm distance in 5 parts.3.Name 2nd part from F as V. It is 20mm 45 and 30mm from F and AB line resp. mm It is first point giving ratio of it’s distances from F and AB 2/3 i.e 20/304. Form more points giving same ratio such F ( focus) (vertex) V as 30/45, 40/60, 50/75 etc.5. Taking 45,60 and 75mm distances from line AB, draw three vertical lines to the right side of it.6. Now with 30, 40 and 50mm distances in compass cut these lines above and below, www.joshikandarp.webs.com with F as center.7. Join these points through V in smooth curve. This is required locus of P. It isan ELLIPSE. B HYPERBOLA - Directrix focus method
- 16. ELLIPSE TANGENT & NORMAL To draw tangent & normal to the curve from a given point ( q )1. JOIN POINT Q TO F1 & F22. BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL3. A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE. p4 C p3 p2 p1 A B O F1 1 2 3 4 F2 www.joshikandarp.webs.com AL RM NO Q TA N GE NT D
- 17. ELLIPSE TANGENT & NORMAL ELLIPSETo draw tangent & normal Ato the curve from a given DIRECTRIXpoint ( Q ) T1. JOIN POINT Q TO F.2. CONSTRUCT 900 ANGLE WITHTHIS LINE AT POINT F (vertex)V3. EXTEND THE LINE TO MEET F ( focus)DIRECTRIX AT T 900 N4. JOIN THIS POINT TO Q ANDEXTEND. THIS IS TANGENT TOELLIPSE FROM Q www.joshikandarp.webs.com5. TO THIS TANGENT DRAW QPERPENDICULAR LINE FROM Q. IT NIS NORMAL TO CURVE. B T
- 18. PARABOLA TANGENT & NORMAL T PARABOLATo draw tangent & normal tothe curve from a given point A(Q)1. JOIN POINT Q TO F.2. CONSTRUCT 900 ANGLE WITHTHIS LINE AT POINT F VERTEX V F3. EXTEND THE LINE TO MEET 900DIRECTRIX AT T ( focus)4. JOIN THIS POINT TO Q ANDEXTEND. THIS IS TANGENT TO THE NCURVE FROM Q www.joshikandarp.webs.com5. TO THIS TANGENT DRAWPERPENDICULAR LINE FROM Q. IT QIS NORMAL TO CURVE. B N T
- 19. HYPERBOLA TANGENT & NORMAL A1. JOIN POINT Q TO F.2. CONSTRUCT 900 ANGLE WITHTHIS LINE AT POINT F3.EXTEND THE LINE TO MEET TDIRECTRIX AT T4. JOIN THIS POINT TO Q AND (vertex V F ( focus) )EXTEND. THIS IS TANGENT TO 900CURVE FROM Q5.TO THIS TANGENT DRAW NERPENDICULAR LINE FROM Q. ITIS NORMAL TO CURVE. www.joshikandarp.webs.com N Q B T

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment