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Curve1 Presentation Transcript

  • 1. ENGINEERING CURVES Part- I {Conic Sections}ELLIPSE PARABOLA HYPERBOLA1.Concentric Circle Method 1.Rectangle Method 1.Rectangular Hyperbola (coordinates given)2.Rectangle Method 2 Method of Tangents ( Triangle Method) 2 Rectangular Hyperbola3.Oblong Method (P-V diagram - Equation given) 3.Basic Locus Method4.Arcs of Circle Method (Directrix – focus) 3.Basic Locus Method (Directrix – focus)5.Rhombus Metho6.Basic Locus Method Methods of Drawing (Directrix – focus) Tangents & Normals To These Curves.
  • 2. ENGINEERING CURVES
  • 3. CONIC SECTIONS ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS BECAUSE THESE CURVES APPEAR ON THE SURFACE OF A CONE WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES. OBSERVE ILLUSTRATIONS GIVEN BELOW.. Ellipse e<1 β α α β β< α β> α Section Plane Section PlaneThrough all the Generators Hyperbola Inclined at an angle Greater than that e>1 of end generator. α β β= α Parabola Section Plane Parallel e=1 to end generator.
  • 4. What is eccentricity ? Conic Directrix A section P N Axis C V D F Focus Vertex B Distance from focus PF VFeccentricity = = = Distance from directrix PN VC
  • 5. COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA: These are the loci of points moving in a plane such that the ratio of it’s distances from a fixed point And a fixed line always remains constant. The Ratio is called ECCENTRICITY. (E) A) For Ellipse E<1 B) For Parabola E=1 C) For Hyperbola E>1 Refer Problem nos. 6. 9 & 12 SECOND DEFINATION OF AN ELLIPSE:- It is a locus of a point moving in a plane such that the SUM of it’s distances from TWO fixed points always remains constant. {And this sum equals to the length of major axis.} These TWO fixed points are FOCUS 1 & FOCUS 2 Refer Problem no.4 Ellipse by Arcs of Circles Method.
  • 6. P C A B F1 F2 AB: Major Axis D CD: Minor Axis PF1+PF2=Constant=AB= Major Axis SECOND DEFINATION OF AN ELLIPSE:- It is a locus of a point moving in a planesuch that the SUM of it’s distances from TWO fixed points always remains constant. {And this sum equals to the length of major axis.} These TWO fixed points are FOCUS 1 & FOCUS 2
  • 7. Problem: Draw an ellipse by general method, given distance of focus from directrix 50 mm andeccentricity 2/3. Also draw normal and tangent on the curve at a point 50 mm from the focus. 1. Draw a vertical line AB of any length as ’ directrix and mark a point C on it. 11 A 10 ’ 2. Draw a horizontal line CD of any length from point C as axis 9’ 3. Mark a point F on line CD at 50 mm from C 8’ 4. Divide CF in 5 equal divisions 7’ 5. Mark V on 2nd division from F 6’ 6. Draw a perpendicular on V and mark a point E 5’ on it at a distance equal to VF 4’ 7. Join CE end extend it 3’ 8. Mark points 1,2,3…on CF beyond V at uniform distance, and draw perpendiculars on each of them 2’ so as to intersect extended CE at 1’,2’,3’... 1’ EC 1 D V F2 3 4 5 6 7 8 9 10 11 90º TAN GEN T AL 90º RM NO B
  • 8. Problem: Draw a parabola by general method, given distance of focus from directrix 50 mm.Also draw normal and tangent on the curve at a point 50 mm from the focus. 1. Draw a vertical line AB of any length as directrix and mark a point C on it. A 2. Draw a horizontal line CD of any length from point C as axis 3. Mark a point F on line CD at 50 mm from C 9’ 8’ 5. Mark V on mid point of CF 7’ 6. Draw a perpendicular on V and mark a point E 6’ on it at a distance equal to VF 5’ 7. Join CE end extend it 4’ 3’ 8. Mark points 1,2,3…on CF beyond V at uniform 2’ distance, and draw perpendiculars on each of them 1’ so as to intersect extended CE at 1’,2’,3’... EC 90º F D V 1 2 3 4 5 6 7 8 9 90º L MA NOR TA NG EN T B
  • 9. Problem 1:- ELLIPSEDraw ellipse by concentric circle method. BY CONCENTRIC CIRCLE METHODTake major axis 150 mm and minor axis 100 mm long.Also draw normal and tangent on the curve at a point 325mm above the major axis 2 4 C Steps: 1 3’ 5 1. Draw both axes as perpendicular bisectors 2’ 4’ of each other & name their ends as shown. P t 2. Taking their intersecting point as a center, 1’ 5’ en ng draw two concentric circles considering both Ta 25mm as respective diameters. No A rm 3. Divide both circles in 12 equal parts & a l B F1 O F2 name as shown. 10’ 6’ 4. From all points of outer circle draw vertical lines downwards and upwards respectively. 9’ 7’ 5.From all points of inner circle draw 10 6 horizontal lines to intersect those vertical 8’ lines. D 6. Mark all intersecting points properly as those are the points on ellipse. 9 7 7. Join all these points along with the ends of 8 both axes in smooth possible curve. It is required ellipse.
  • 10. Steps: ELLIPSE BY RECTANGLE METHOD1 Draw a rectangle taking majorand minor axes as sides. Problem 22. In this rectangle draw both Draw ellipse by Rectangle method.Take major axis 100 mm andaxes as perpendicular bisectors minor axis 70 mm long. Also draw a normal and a tangent on theof each other..3. For construction, select upper curve at a point 25 mm above the major axis.left part of rectangle. Dividevertical small side and horizontallong side into same number of D 4’equal parts.( here divided in four Tanparts) gen 3’ t4. Name those as shown..5. Now join all vertical points 2’ 25mm1’,2’,3’,4’, to the upper end ofminor axis. And all horizontal 1’ l rmapoints i.e.1,2,3,4 to the lower end A B Noof minor axis. 2 3 4O 16. Then extend C-1 line upto D-1’ and mark that point. Similarlyextend C-2, C-3, C-4 lines up toD-2’, D-3’, & D-4’ lines.7. Mark all these points properlyand join all along with ends Aand D in smooth possible curve.Points in the remaining three Cquadrants can be marked usingprincipal of symmetry. Then joinall the points so obtained. It isrequired ellipse.
  • 11. PROBLEM 4. ELLIPSEMAJOR AXIS AB & MINOR AXIS CD ARE BY ARCS OF CIRCLE METHOD100 AMD 70MM LONG RESPECTIVELY.DRAW ELLIPSE BY ARCS OF CIRLESMETHOD. As per the definition Ellipse is locus of point P moving in a plane such that the SUM of it’s distances from two fixedSTEPS: points (F1 & F2) remains constant and equals to the length1.Draw both axes as usual.Name the of major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB) ends & intersecting point2.Taking AO distance I.e.half major axis, from C, mark F1 & F2 On AB . P4 C P4 ( focus 1 and 2.) P3 P33.On line F1- O taking any distance, P2 P2 P1 mark points 1,2,3, & 4 P14.Taking F1 center, with distance A-1 draw an arc above AB and taking F2 center, with B-1 distance cut this arc. Name the point p1 A B O F25.Repeat this step with same centers but F1 1 2 3 4 taking now A-2 & B-2 distances for drawing arcs. Name the point p2 P1 P16.Similarly get all other P points. With same steps positions of P can be P2 P2 located below AB. P3 P37.Join all points by smooth curve to get P4 P4 an ellipse/ D
  • 12. ELLIPSEProblem 13: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) 1. JOIN POINT Q TO F1 & F2 2. BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL 3. 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 ALM NOR Q TAN GE NT D
  • 13. ELLIPSE Problem 3:- BY OBLONG METHOD Draw ellipse by Oblong method. Draw a parallelogram of 100 mm and 70 mm long sides with included angle of 750.Inscribe Ellipse in it. STEPS ARE SIMILAR TO THE PREVIOUS CASE (RECTANGLE METHOD) ONLY IN PLACE OF RECTANGLE, HERE IS A PARALLELOGRAM. D 4 4 3 3 2 2 1 1A 1 2 3 4 3 2 1 B C
  • 14. PROBLEM 7: A BALL THROWN IN AIR ATTAINS 100 M HIEGHT PARABOLA AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND. RECTANGLE METHOD Draw the path of the ball (projectile)- Scale 1cm = 10m.STEPS: 61.Draw rectangle of above size and divide it in two equal vertical parts2.Consider left part for construction. 5 Divide height and length in equal number of parts and name those 1,2,3,4,5& 6 43.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle4.Similarly draw upward vertical lines from horizontal1,2,3,4,5 3 And wherever these lines intersect previously drawn inclined lines in sequence Mark those points and 2 further join in smooth possible curve.5.Repeat the construction on right side rectangle also.Join all in sequence. 1 This locus is Parabola.. 1’ 2’ 3’ 4’ 5’ 6’
  • 15. Draw a parabola by tangent method given base 7.5m and axis 4.5m Take scale 1cm = 0.5m O 10 9 1’ 8 2’ 7 3’ 4.5m 6 4’ 5 F 5’ 4 6’ 3 7’ 4.5m 2 8’ 1 9’ 10’ E B A 7.5m
  • 16. Problem 51: A fountain jet discharges water from ground level at an inclination of 45º tothe ground. The jet travels a horizontal distance of 7.5m from the point of discharge andfalls on the ground. Trace the path of the jet. Name the curve.As the jet will be a projectile so its path will be parabolic. The angle of jet with the ground is the angle of tangenton the curve at the point of discharge. First we will consider a scale to accommodate 7.5 m on the ground. Thatcan be done by considering 1cm= 0.5 m. O 5 4 1’ 3 2’ 2 3’ 1 4’ 5’ 45º 45º B A 7.5m
  • 17. Problem No.10: Point P is 40 mm and 30 mm from horizontal HYPERBOLAand vertical axes respectively.Draw Hyperbola through it. THROUGH A POINT OF KNOWN CO-ORDINATES Solution Steps: 1) Extend horizontal line from P to right side. 1’ 2) Extend vertical line from P upward. 3) On horizontal line from P, mark some points taking any distance and name them after P-1, 2,3,4 etc. 4) Join 1-2-3-4 points to pole O. Let them cut 2’ part [P-B] also at 1,2,3,4 points. 5) From horizontal 1,2,3,4 draw vertical 1 2 P 3 4 5 lines downwards and 6) From vertical 1,2,3,4 points [from P-B] draw 3’ horizontal lines. 7) Line from 1 40 mm 4’ horizontal and line from 1 vertical will meet at 5’ P1.Similarly mark P2, P3, P4 points. O 8) Repeat the procedure by marking four points 30 mm on upward vertical line from P and joining all those to pole O. Name this points P6, P7, P8 etc. and join them by smooth
  • 18. Arc of circle MethodProblem 14: Two points A and B are 50 mm apart. A point P moves in a plane in such away that the difference of its distance from A and B is always constant and equal to 20mm. Draw the locus of point P. Draw a line and mark two points A & B on it at a distance of 50 mm. Mark O as mid point of AB. Mark two points V1 and V2 at 10 mm on either side of O. Mark points 1, 2,3 on the right of Bat any distances. o As per the definition Hyperbola is locus of point P moving in a plane such that the difference of it’s V1 V2 1 2 3 A B distances from two fixed points (F1 & F2) remains constant and equals to the length of transverse axis V1 V2. 10 10 Take V11 as radius and A as centre and 50 draw an arc on the right side of A. Take V21 as radius and B as centre and draw an arc on the left side of B so as to intersect the previous arc. Repeat the step with V12, V22 as radii and V13, V23 as radii respectively. Repeat the same steps on the other side to draw the second hyperbola.
  • 19. ELLIPSEProblem 14: TANGENT & NORMALTO DRAW TANGENT & NORMAL TO THE CURVE ELLIPSE FROM A GIVEN POINT ( Q ) A DIRECTRIX1.JOIN POINT Q TO F. T2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO ELLIPSE FROM Q (vertex) V5.TO THIS TANGENT DRAW PERPENDICULAR F ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. 900 N Q N B T
  • 20. PARABOLA Problem 15: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE T PARABOLA FROM A GIVEN POINT ( Q ) A1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS VERTEX V TANGENT TO THE CURVE FROM Q 900 F5.TO THIS TANGENT DRAW PERPENDICULAR ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. N Q B N T
  • 21. HYPERBOLA Problem 16 TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) A1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F T3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO CURVE FROM Q (vertex) V F ( focus)5.TO THIS TANGENT DRAW PERPENDICULAR 900 LINE FROM Q. IT IS NORMAL TO CURVE. N N Q B T