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- 1. Engineering CurvesAshish Vishnu ShelkeLecturer,Mechanical Engineering DepartmentPCP,Pune:+919960818069 1
- 2. 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
- 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 Section Plane ola Section PlaneThrough Generators Hyperbola rab Parallel to Axis. Pa Section Plane Parallel to end generator. 3
- 4. 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. 4
- 5. ELLIPSE BY CONCENTRIC CIRCLE METHOD Problem 1 :- Draw ellipse by concentric circle method.Take major axis 100 mm and minor axis 70 mm long. 3 2 4 Steps: 1. Draw both axes as perpendicular bisectors C of each other & name their ends as shown. 2. Taking their intersecting point as a center, 1 5 3 2 4 draw two concentric circles considering both as respective diameters. 1 5 3. Divide both circles in 12 equal parts & name as shown. A 4. From all points of outer circle draw vertical B lines downwards and upwards respectively. 5.From all points of inner circle draw 10 6 horizontal lines to intersect those vertical lines. 10 9 7 6 6. Mark all intersecting points properly as 8 those are the points on ellipse. D 7. Join all these points along with the ends of both axes in smooth possible curve. It is 9 7 required ellipse. 8 5
- 6. Steps: ELLIPSE BY RECTANGLE METHOD1 Draw a rectangle taking majorand minor axes as sides.2. In this rectangle draw both Problem 2axes as perpendicular bisectors Draw ellipse by Rectangle method.of each other..3. For construction, select upper Take major axis 100 mm and minor axis 70 mm long.left part of rectangle. Dividevertical small side and horizontallong side into same number of D 4 4equal parts.( here divided in fourparts) 3 34. Name those as shown..5. Now join all vertical points 2 21,2,3,4, to the upper end of minoraxis. And all horizontal points 1 1i.e.1,2,3,4 to the lower end ofminor axis. A 1 2 3 4 3 2 1 B6. Then extend C-1 line upto D-1and 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.Do similar construction in right Cside part.along with lower half ofthe rectangle.Join all points insmooth curve. 6It is required ellipse.
- 7. 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 7 C
- 8. ELLIPSE BY ARCS OF CIRCLE METHOD Ellipse can be defined as a curve traced by point moving in the same plane such that the sum of its distances from the two fixed points is always constant. From the fig, AB=Major Axis CD= Minor axis The fixed points(Focii) are equidistant from centre ‘O’ Then by definition F1C+F2C=F1E+F2E=F1A+F2A Major axis, AB=F2B+F1F2+F1A=F2B+F2A Since, F2B=F1A F1A=F2A=Length of Major Axis Therefore, F1C+F2C=F1E+F2E=Length of Major Axis Hence we can say that sum of distance of any point on the curve from two fixed points is equal to the major axis But F1C=F2C Therefore F1C= Major Axis/2 8
- 9. PROBLEM 4.MAJOR AXIS AB & MINOR AXIS CD ARE100 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 ( focus 1 and 2.) p33.On line F1- O taking any distance, p2 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 O5.Repeat this step with same centers but F1 1 2 3 4 F2 taking now A-2 & B-2 distances for drawing arcs. Name the point p26.Similarly get all other P points. With same steps positions of P can be located below AB.7.Join all points by smooth curve to get an ellipse/ D 9
- 10. : ELLIPSE 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 10
- 11. PROBLEM 5. ELLIPSEDRAW RHOMBUS OF 100 MM & 70 MM LONG BY RHOMBUS METHODDIAGONALS AND INSCRIBE AN ELLIPSE IN IT.STEPS: 21. Draw rhombus of given dimensions.2. Mark mid points of all sides & name Those A,B,C,& D3. Join these points to the ends of A B smaller diagonals.4. Mark points 1,2,3,4 as four centers.5. Taking 1 as center and 1-A 3 4 radius draw an arc AB.6. Take 2 as center draw an arc CD.7. Similarly taking 3 & 4 as centers and 3-D radius draw arcs DA & BC. D C 1 11
- 12. PROBLEM 6:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANESUCH THAT THE RATIO OF IT’S DISTANCES FROM F AND LINE AB REMAINS CONSTANT ELLIPSEAND EQUALS TO 2/3 DRAW LOCUS OF POINT P. { ECCENTRICITY = 2/3 } DIRECTRIX-FOCUS METHOD ELLIPSE A STEPS: 1 .Draw a vertical line AB and point F DIRECTRIX 50 mm from it. 2 .Divide 50 mm distance in 5 parts. 45mm 3 .Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. 30mm It is first point giving ratio of it’s distances from F and AB 2/3 i.e 20/30 4 Form more points giving same ratio such as 30/45, 40/60, 50/75 etc. (vertex) V 5.Taking 45,60 and 75mm distances from F ( focus) 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, with F as center. 7. Join these points through V in smooth curve. This is required locus of P.It is an ELLIPSE. B 12
- 13. 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 13 T
- 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)-STEPS: 6 61.Draw rectangle of above size and divide it in two equal vertical parts2.Consider left part for construction. 5 5 Divide height and length in equal number of parts and name those 1,2,3,4,5& 6 4 43.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle4.Similarly draw upward vertical 3 3 lines from horizontal1,2,3,4,5 And wherever these lines intersect previously drawn inclined lines in sequence Mark those points and 2 2 further join in smooth possible curve.5.Repeat the construction on right side rectangle also.Join all in sequence. 1 1 This locus is Parabola.. 1 2 3 4 5 6 5 4 3 2 1 14
- 15. Problem no.8: Draw an isosceles triangle of 100 mm long base and PARABOLA110 mm long altitude.Inscribe a parabola in it by method of tangents. METHOD OF TANGENTS Solution Steps: C 1. Construct triangle as per the given 14 1 dimensions. 2. Divide it’s both sides in to same no.of 13 2 equal parts. 12 3 3. Name the parts in ascending and 11 descending manner, as shown. 4 10 4. Join 1-1, 2-2,3-3 and so on. 5 5. Draw the curve as shown i.e.tangent to 9 6 all these lines. The above all lines being 8 tangents to the curve, it is called method 7 of tangents. 7 8 6 9 5 10 4 11 3 12 2 13 1 14 A B 15
- 16. PROBLEM 9: Point F is 50 mm from a vertical straight line AB. PARABOLADraw locus of point P, moving in a plane such that DIRECTRIX-FOCUS METHODit always remains equidistant from point F and line AB. PARABOLASOLUTION STEPS:1.Locate center of line, perpendicular to A AB from point F. This will be initial point P and also the vertex.2.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from P1those draw lines parallel to AB.3.Mark 5 mm distance to its left of P and (VERTEX) V name it 1. F O 1 2 3 44.Take O-1 distance as radius and F as center draw an arc ( focus) cutting first parallel line to AB. Name upper point P1 and lower point P2. P2(FP1=O1)5.Similarly repeat this process by taking again 5mm to right and left and locate P3 P4 . B6.Join all these points in smooth curve. It will be the locus of P equidistance from line AB and fixed point F. 16
- 17. 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 17
- 18. 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 18 T
- 19. Hyperbola: Hyperbola is the locus of point which moves in a plane so that the ratio of its distances from a focu and directrix is constant and greater than unity. Applications: Shape of overhear tank, shape of cooling towers, shape of pv diagram etc… Methods of construction of hyperbola:1) Directrix focus method2) Rectangular Hyperbola 19
- 20. Assignment1 An artificial satellite is orbiting around the earth. The major axis of the orbit is 40000km and minor axis is 30000Km.Draw the orbit of the satellite and show the position of the earth center assuming that earth is at the focus. Draw tangent and normal when the satellite is 10000Km from the earth centre. (Dec2008) Draw an ellipse with major axis 150mm and minor axis110mm.Left side of the minor axis is to be draw by concentric circle method and right side of minor axis by rectangle method .Draw tangent and normal to the ellipse at 50mm radius from center of ellipse.(May 2009) Draw a parallelogram of 160mm and 120mm sides with included angle of 1200.Inscribe an ellipse within the parallelogram. Determine the major and minor axis of the ellipse .Draw the tangent and normal to the ellipse at a point 20mm above the major axis and at the left side of center . (Oct 2009) Draw an Ellipse with the major axis 160mm and minor axis 120mm.The portion on the left side of the minor axis is to be drawn by concentric circle method and on the right side of minor axis by rectangle method. Draw tangent and normal to the ellipse at a point 70 mm Distance from the center of ellipse. (May 2009) 20
- 21. Assignment1 Draw a parabola that has distance of 50 mm between the focus and the directrix. Draw normal and tangent to the parabola at a point 30 mm from the focus. Draw a parabola within a parallelogram ABCD of sides AB=75mm, BC =85mm and angle ABC=750 Using tangent method draw a parabola with 60 mm base length and 25 mm axis length. Two asymptotes OX and OY are at 750 angles with each other. Point P is 30 mm and 40mm away from OX and OY respectively. Draw the hyperbola passing through point P taking at least 10 points. Draw tangent and normal at point von hyperbola to away from OX. (Oct 2009) 21
- 22. ENGINEERING CURVES Part-II (Point undergoing two types of displacements) INVOLUTE CYCLOID SPIRAL HELIX1. Involute of a circle 1. General Cycloid 1. Spiral of 1. On Cylinder a)String Length = πD One Convolution. 2. Trochoid 2. On a Cone b)String Length > πD ( superior) 2. Spiral of 3. Trochoid Two Convolutions. c)String Length < πD ( Inferior) 4. Epi-Cycloid2. Pole having Composite shape. 5. Hypo-Cycloid3. Rod Rolling over a Semicircular Pole. AND Methods of Drawing Tangents & Normals To These Curves. 22
- 23. DEFINITIONSCYCLOID:IT IS A LOCUS OF A POINT ON THE SUPERIORTROCHOID:PERIPHERY OF A CIRCLE WHICH IF THE POINT IN THEROLLS ON A STRAIGHT LINE PATH. DEFINATION OF CYCLOID IS OUTSIDE THE CIRCLEINVOLUTE: INFERIOR TROCHOID.:IT IS A LOCUS OF A FREE END OF A STRINGIF IT IS INSIDE THE CIRCLEWHEN IT IS WOUND ROUND A CIRCULAR POLE EPI-CYCLOID IF THE CIRCLE IS ROLLING ONSPIRAL: ANOTHER CIRCLE FROMIT IS A CURVE GENERATED BY A POINT OUTSIDEWHICH REVOLVES AROUND A FIXED POINT HYPO-CYCLOID.AND AT THE SAME MOVES TOWARDS IT. IF THE CIRCLE IS ROLLING FROM INSIDE THE OTHER CIRCLE,HELIX:IT IS A CURVE GENERATED BY A POINT WHICHMOVES AROUND THE SURFACE OF A RIGHT CIRCULARCYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIALDIRECTION 23AT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OFROTATION.
- 24. Involute An involute is a curve traced by a point as it unwindsfrom around a circle or polygon.The concerned circle orpolygon is called as evolute. 24
- 25. Draw a involute of Equilatral triangle of 20 mm SideSteps: Draw the equilatral triangle of 20 mm side Imagine a piece of thread which is already wound on the triangle is being unwound Produce the line BA ,AC and CB. With A as Centre and AB as radius Draw an arc to intersect CB at P1 Again B as centre and BP1 as Radius draw an arc to intersect AC at P2 Lastly With C as centre and CP2 as radius to intersect BA at P3 Join Point A,P1,P2,P3 by smooth curve to get required involute 25
- 26. Problem : Draw Involute of a circle. INVOLUTE OF A CIRCLE String length is equal to the circumference of circle.Solution Steps:1) Point or end P of string AP isexactly πD distance away from A.Means if this string is wound roundthe circle, it will completely cover P2given circle. B will meet A afterwinding.2) Divide πD (AP) distance into 8 P3number of equal parts. P13) Divide circle also into 8 number 2 to pof equal parts. 34) Name after A, 1, 2, 3, 4, etc. up to pto 8 on πD line AP as well as on pcircle (in anticlockwise direction). o 1t5) To radius C-1, C-2, C-3 up to C-8draw tangents (from 1,2,3,4,etc to 4 to pcircle). P4 46) Take distance 1 to P in compass 3and mark it on tangent from point 1 5on circle (means one division less 2than distance AP). 6 p o7) Name this point P1 5t 18) Take 2-B distance in compass 7 A 8 6 to pand mark it on the tangent from 7 to Ppoint 2. Name it point P2. P5 p P8 1 2 3 4 5 6 7 89) Similarly take 3 to P, 4 to P, 5 to P7P up to 7 to P distance in compass P6 πand mark on respective tangentsand locate P3, P4, P5 up to P8 (i.e. DA) points and join them in smoothcurve it is an INVOLUTE of a given 26circle.
- 27. INVOLUTE OF A CIRCLEProblem: Draw Involute of a circle. String length MORE than πDString length is MORE than the circumference of circle.Solution Steps: P2In this case string length is morethan Π D. But remember!Whatever may be the length of P3 P1string, mark Π D distance 2 to phorizontal i.e.along the stringand divide it in 8 number of 3 toequal parts, and not any other p pdistance. Rest all steps are same o 1tas previous INVOLUTE. Drawthe curve completely. 4 to p P4 4 3 5 2 p o 5t 6 1 P5 7 8 7 p8 1 P 6 to p to p 2 3 4 5 6 7 8 P7 165 mm P6 (more than πD) πD 27
- 28. Problem : Draw Involute of a circle. INVOLUTE OF A CIRCLEString length is LESS than the circumference of circle. String length LESS than πDSolution Steps: P2In this case string length is Lessthan Π D. But remember!Whatever may be the length of P3 P1string, mark Π D distancehorizontal i.e.along the stringand divide it in 8 number of 2 to p 3 toequal parts, and not any other pdistance. Rest all steps are sameas previous INVOLUTE. Draw op 1tthe curve completely. 4 to p P4 4 3 5 2 p o 6 5t 1 6 to p P5 7 to 7 P p 8 P7 1 2 3 4 5 6 7 8 P6 150 mm (Less than πD) πD 28
- 29. PROBLEM : A POLE IS OF A SHAPE OF HALF HEXABON AND SEMICIRCLE. ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER INVOLUTE DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY. OF (Take hex 30 mm sides and semicircle of 60 mm diameter.) COMPOSIT SHAPED POLESOLUTION STEPS:1)Draw pole shape as perdimensions. P12)Divide semicircle in 4 partsand name those along with Pcorners of hexagon. P23)Calculate perimeter length.4)Show it as string AP.On this line mark 30mm from 1 to PA5)Mark and name it 1 2 toMark πD/2 distance on it from P oP1 AtAnd dividing it in 4 partsname 2,3,4,5.6)Mark point 6 on line 30 mm P 3from 5 3 to P 37)Now draw tangents from all 4 2points of poleand proper lengths as done in 5 1all previous oP involute’s problems and A 4tcomplete the curve. 6 5 to P 1 2 3 4 5 6 P 6t oP P4 πD/2 P6 P5 29
- 30. Archemedian Spiral It is curve generated by point moving uniformly along a straight line while a line swings or rotates around a fixed point. The point about which point swings is called pole, each complete revolution is termed as convolution.Application Springs in watch mechanism Profile of cam for automobile Screw threads 30
- 31. Problem 27: Draw a spiral of one convolution. Take distance PO 40 mm. SPIRAL IMPORTANT APPROACH FOR CONSTRUCTION! FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2 P2Solution Steps 3 1 P11. With PO radius draw a circle and divide it in EIGHT parts. P3 Name those 1,2,3,4, etc. up to 82 .Similarly divided line PO also in EIGHT parts and name those 4 P4 O P 1,2,3,-- as shown. 7 6 5 4 3 2 13. Take o-1 distance from op line P7 and draw an arc up to O1 radius P5 P6 vector. Name the point P14. Similarly mark points P2, P3, P4 up to P8 5 7 And join those in a smooth curve. It is a SPIRAL of one convolution. 31 6
- 32. Problem 28 SPIRALPoint P is 80 mm from point O. It starts moving towards O and reaches it in two ofrevolutions around.it Draw locus of point P (To draw a Spiral of TWO convolutions). two convolutions IMPORTANT APPROACH FOR CONSTRUCTION! FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2,10 P2 3,11 P1 1,9 SOLUTION STEPS: P3 Total angular displacement here P10 is two revolutions And P9 Total Linear displacement here P11 is distance PO. 16 13 10 8 7 6 5 4 3 2 1 P Just divide both in same parts i.e. 4,12 P4 P8 8,16 P12 Circle in EIGHT parts. P15 ( means total angular displacement P13 P14 in SIXTEEN parts) Divide PO also in SIXTEEN parts. P7 Rest steps are similar to the previous P5 problem. P6 5,13 7,15 32 6,14
- 33. Spiral. Method of Drawing Tangent & Normal SPIRAL (ONE CONVOLUSION.) 2 nt ge No n Ta rm P2 al 3 1 Difference in length of any radius vectors Q P1 Constant of the Curve = Angle between the corresponding radius vector in radian. P3 OP – OP2 OP – OP2 = = π/2 1.574 P4 O P = 3.185 m.m. 7 6 5 4 3 2 1 P7 STEPS: *DRAW SPIRAL AS USUAL. P5 P6 DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE CONSTANT OF CURVE CALCULATED ABOVE. * LOCATE POINT Q AS DISCRIBED IN PROBLEM AND 5 7 THROUGH IT DRAW A TANGENTTO THIS SMALLER CIRCLE.THIS IS A NORMAL TO THE SPIRAL. *DRAW A LINE AT RIGHT ANGLE 6 *TO THIS LINE FROM Q. IT WILL BE TANGENT TO CYCLOID. 33
- 34. Cycloid A cycloid is a curve generated by a point on the circumference of a circle as the circle rolls along a straight line without sliping The rolling circle is called generating circle and the line along which it rolls is called base line or directing line. 34
- 35. PROBLEM 22: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE CYCLOIDWHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm p4 4 p3 p5 3 5 C p2 C1 C2 C3 C4 C5 C6 C7 p6 C 8 2 6 p1 1 p7 7 P p8 πD Solution Steps: 1) From center C draw a horizontal line equal to πD distance. 2) Divide πD distance into 8 number of equal parts and name them C1, C2, C3__ etc. 3) Divide the circle also into 8 number of equal parts and in clock wise direction, after P name 1, 2, 3 up to 8. 4) From all these points on circle draw horizontal lines. (parallel to locus of C) 5) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P. 6) Repeat this procedure from C2, C3, C4 upto C8 as centers. Mark points P2, P3, P4, P5 up to P8 on the horizontal lines drawn from 2, 3, 4, 5, 6, 7 respectively. 7) Join all these points by curve. It is Cycloid. 35
- 36. PROBLEM 23: DRAW LOCUS OF A POINT , 5 MM AWAY FROM THE PERIPHERY OF A SUPERIOR TROCHOID CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm 4 p4 p3 p5 3 5 p2 C C1 C C3 C4 C5 C6 C7 C8 p 6 2 6 2 p7 1 p1 7 P πD p8Solution Steps:1) Draw circle of given diameter and draw a horizontal line from it’s center C of length Π D and divide it in 8 number of equal parts and name them C1, C2, C3, up to C8.2) Draw circle by CP radius, as in this case CP is larger than radius of circle.3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number of equal parts and drawing lines from all these points parallel to locus of C and taking CP radius wit 36 different positions of C as centers, cut these lines and get different positions of P and join4) This curve is called Superior Trochoid.
- 37. PROBLEM 24: DRAW LOCUS OF A POINT , 5 MM INSIDE THE PERIPHERY OF A INFERIOR TROCHOID CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm p4 4 p3 p5 3 5 p2 C C1 C2 C3 C4 C5 C6 C7 p6 C8 2 6 p1 p7 1 7 P p8 πDSolution Steps:1) Draw circle of given diameter and draw a horizontal line from it’s center C of length Π D and divide it in 8 number of equal parts and name them C1, C2, C3, up to C8.2) Draw circle by CP radius, as in this case CP is SHORTER than radius of circle.3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number of equal parts and drawing lines from all these points parallel to locus of C and taking CP radius with different positions of C as centers, cut these lines and get different positions of P and join 37 those in curvature.4) This curve is called Inferior Trochoid.
- 38. Epi cycloid An epicycloid is a curve generated by a point on thecircumference of a circle which rolls on the outside of anothercircle without sliding or slipping. The rolling circle is called generating circle and theoutside circle on which it rolls is called the directing circle or thebase circle. 38
- 39. PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLEWHICH ROLLS ON A CURVED PATH. Take diameter of rolling Circle 50 mm EPI CYCLOID :And radius of directing circle i.e. curved path, 75 mm.Solution Steps:1) When smaller circle will roll onlarger circle for one revolution it willcover Π D distance on arc and it will bedecided by included arc angle θ.2) Calculate θ by formula θ = (r/R) x Generating/3600. Rolling Circle3) Construct angle θ with radius OC 4 5and draw an arc by taking O as center C2 C3 C1 C4OC as radius and form sector of angle 3 6θ. C C 54) Divide this sector into 8 number of 7equal angular parts. And from C 2 C6onward name them C1, C2, C3 up toC8. 1 P r = CP C75) Divide smaller circle (Generatingcircle) also in 8 number of equal parts.And next to P in clockwise direction Directing Circlename those 1, 2, 3, up to 8. R C 86) With O as center, O-1 as radiusdraw an arc in the sector. Take O-2, O- = r 3600 +3, O-4, O-5 up to O-8 distances with Rcenter O, draw all concentric arcs in Osector. Take fixed distance C-P incompass, C1 center, cut arc of 1 at P1.Repeat procedure and locate P2, P3,P4, P5 unto P8 (as in cycloid) and join 39them by smooth curve. This is EPI –CYCLOID.
- 40. Hypocycloid A hypocycloid is a curve generated by a point on the circumference of a circle which rolls on the inside of another circle without sliding or slipping. The rolling circle is called generating circle/hypo circle and the inside circle on which it rolls is called the directing circle or the base circle. 40
- 41. PROBLEM 26: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLEWHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter of HYPO CYCLOIDrolling circle 50 mm and radius of directing circle (curved path) 75 mm. Solution Steps: 1) Smaller circle is rolling here, inside the larger circle. It has to rotate anticlockwise to move P 7 ahead. 2) Same steps should be P1 6 taken as in case of EPI – CYCLOID. Only change is 1 P2 C2 C1 C3 in numbering direction of 8 C4 number of equal parts on C C P3 5 5 the smaller circle. 2 C 3) From next to P in 6 anticlockwise direction, 4 P4 C name 1,2,3,4,5,6,7,8. 3 7 4) Further all steps are P5 P8 that of epi – cycloid. This P6 P7 is called C8 HYPO – CYCLOID. r 3600 = R + O OC = R ( Radius of Directing Circle) CP = r (Radius of Generating Circle) 41
- 42. HELIX (UPON A CYLINDER)PROBLEM: Draw a helix of one convolution, upon a cylinder. P8Given 80 mm pitch and 50 mm diameter of a cylinder. 8(The axial advance during one complete revolution is called P7The pitch of the helix) 7 P6 6 P5SOLUTION: 5Draw projections of a cylinder.Divide circle and axis in to same no. of equal parts. ( 8 ) 4 P4Name those as shown. 3Mark initial position of point ‘P’ P3Mark various positions of P as shown in animation. 2 P2Join all points by smooth possible curve.Make upper half dotted, as it is going behind the solid 1 P1and hence will not be seen from front side. P 6 7 5 P 4 1 3 2 42
- 43. HELIXPROBLEM: Draw a helix of one convolution, upon a cone, 8 P8 (UPON A CONE)diameter of base 70 mm, axis 90 mm and 90 mm pitch.(The axial advance during one complete revolution is called 7 P7The pitch of the helix) 6 P6 P5 SOLUTION: 5 Draw projections of a cone Divide circle and axis in to same no. of equal parts. ( 8 ) 4 P4 Name those as shown. Mark initial position of point ‘P’ 3 P3 Mark various positions of P as shown in animation. Join all points by smooth possible curve. 2 P2 Make upper half dotted, as it is going behind the solid 1 and hence will not be seen from front side. P1 X P Y 6 7 5 P6 P5 P7 P4 P 4 P8 P1 P3 1 3 P2 43 2
- 44. STEPS: InvoluteDRAW INVOLUTE AS USUAL. Method of DrawingMARK POINT Q ON IT AS DIRECTED. Tangent & NormalJOIN Q TO THE CENTER OF CIRCLE C.CONSIDERING CQ DIAMETER, DRAWA SEMICIRCLE AS SHOWN. INVOLUTE OF A CIRCLE l maMARK POINT OF INTERSECTION OF r NoTHIS SEMICIRCLE AND POLE CIRCLEAND JOIN IT TO Q. QTHIS WILL BE NORMAL TO INVOLUTE. Ta ngDRAW A LINE AT RIGHT ANGLE TO enTHIS LINE FROM Q. tIT WILL BE TANGENT TO INVOLUTE. 4 3 5 C 2 6 1 7 8 P P8 1 2 3 4 5 6 7 8 π D 44
- 45. STEPS:DRAW CYCLOID AS USUAL. CYCLOIDMARK POINT Q ON IT AS DIRECTED. Method of DrawingWITH CP DISTANCE, FROM Q. CUT THE Tangent & NormalPOINT ON LOCUS OF C AND JOIN IT TO Q.FROM THIS POINT DROP A PERPENDICULARON GROUND LINE AND NAME IT NJOIN N WITH Q.THIS WILL BE NORMAL TOCYCLOID.DRAW A LINE AT RIGHT ANGLE TOTHIS LINE FROM Q. al No r mIT WILL BE TANGENT TO CYCLOID. CYCLOID Q Tang e nt CP C C1 C2 C3 C4 C5 C6 C7 C8 P N 45 πD
- 46. LOCUS It is a path traced out by a point moving in a plane, in a particular manner, for one cycle of operation.The cases are classified in THREE categories for easy understanding. A} Basic Locus Cases. B} Oscillating Link…… C} Rotating Link………Basic Locus Cases:Here some geometrical objects like point, line, circle will be described with there relativePositions. Then one point will be allowed to move in a plane maintaining specific relationwith above objects. And studying situation carefully you will be asked to draw it’s locus.Oscillating & Rotating Link:Here a link oscillating from one end or rotating around it’s center will be described.Then a point will be allowed to slide along the link in specific manner. And now studyingthe situation carefully you will be asked to draw it’s locus.STUDY TEN CASES GIVEN ON NEXT PAGES 46
- 47. Basic Locus Cases: PROBLEM 1.: Point F is 50 mm from a vertical straight line AB. Draw locus of point P, moving in a plane such that it always remains equidistant from point F and line AB. P7 A P5SOLUTION STEPS:1.Locate center of line, perpendicular to P3 AB from point F. This will be initial point P. P12.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from those draw lines parallel to AB.3.Mark 5 mm distance to its left of P and p name it 1. 1 2 3 4 F 4 3 2 14.Take F-1 distance as radius and F as center draw an arc cutting first parallel line to AB. Name upper point P1 and lower point P2. P25.Similarly repeat this process by taking again 5mm to right and left and locate P4 P3 P4 .6.Join all these points in smooth curve. P6 B P8 It will be the locus of P equidistance from line AB and fixed point F. 47
- 48. Basic Locus Cases:PROBLEM 2 :A circle of 50 mm diameter has it’s center 75 mm from a verticalline AB.. Draw locus of point P, moving in a plane such thatit always remains equidistant from given circle and line AB. P7 P5 ASOLUTION STEPS: P31.Locate center of line, perpendicular to 50 D AB from the periphery of circle. This will be initial point P. P12.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from those draw lines parallel to AB.3.Mark 5 mm distance to its left of P and p name it 1,2,3,4. C 4 3 2 1 1 2 3 44.Take C-1 distance as radius and C as center draw an arc cutting first parallel line to AB. Name upper point P1 and lower point P2. P25.Similarly repeat this process by taking again 5mm to right and left and locate P3 P4 . P46.Join all these points in smooth curve. B P6 It will be the locus of P equidistance P8 from line AB and given circle. 75 mm 48
- 49. PROBLEM 3 : Basic Locus Cases: Center of a circle of 30 mm diameter is 90 mm away from center of another circle of 60 mm diameter. Draw locus of point P, moving in a plane such that it always remains equidistant from given two circles.SOLUTION STEPS:1.Locate center of line,joining two 60 Dcenters but part in between periphery P7of two circles.Name it P. This will be P5 30 Dinitial point P. P32.Mark 5 mm distance to its rightside, name those points 1,2,3,4 and P1from those draw arcs from C1As center.3. Mark 5 mm distance to its right pside, name those points 1,2,3,4 and C1 C2 4 3 2 1 1 2 3 4from those draw arcs from C2 Ascenter. P24.Mark various positions of P as perprevious problems and name those P4similarly. P65.Join all these points in smoothcurve. P8 It will be the locus of Pequidistance from given twocircles. 95 mm 49
- 50. Basic Locus Cases: Problem 4:In the given situation there are two circles of different diameters and one inclined line AB, as shown. Draw one circle touching these three objects. 60 DSolution Steps: 30 D1) Here consider two pairs,one is a case of two circleswith centres C1 and C2 anddraw locus of point Pequidistance from them. C1 C(As per solution of case D 1 C2 350above). 2) Consider second casethat of fixed circle (C1) andfixed line AB and drawlocus of point Pequidistance from them.(as per solution of case Babove). 3) Locate the point wherethese two loci intersecteach other. Name it x. Itwill be the pointequidistance from giventwo circles and line AB. 4) Take x as centre and itsperpendicular distance onAB as radius, draw a circlewhich will touch given twocircles and line AB. 50
- 51. Basic Locus Cases:Problem 5:-Two points A and B are 100 mm apart. There is a point P, moving in a plane such that the difference ofit’s distances from A and B always remains constant and equals to 40 mm. Draw locus of point P. Solution Steps: 1.Locate A & B points 100 p7 mm apart. p5 2.Locate point P on AB line, p3 70 mm from A and 30 mm from B p1 As PA-PB=40 ( AB = 100 mm ) 3.On both sides of P mark P A B points 5 4 3 2 1 1 2 3 4 mm apart. Name those 1,2,3,4 as usual. p2 4.Now similar to steps of p4 Problem 2, p6 Draw different arcs taking A & B centers p8 and A-1, B-1, A-2, B-2 etc as radius. 70 mm 30 mm 5. Mark various positions of p i.e. and join them in smooth possible curve. It will be locus of P 51
- 52. Problem 6:-Two points A and B are 100 mm apart. There is a FORK & point P, moving in a plane such that the difference of it’sdistances from A and B always remains constant and equals to 40 mm. SLIDER A Draw locus of point P. MSolution Steps: p M11) Mark lower most p1 M2position of M on extension C p2of AB (downward) by N3 N5 M3taking distance MN (40 N6 p3 N2mm) from point B N4 p4 M4 N1(because N can not go N7 N p5 N8 9beyond B ). N10 90 0 M5 p2) Divide line (M initial N 6 M6 N11 and M lower most ) into p7 60 0eight to ten parts and N12 p8 B M7 N13mark them M1, M2, M3 upto the last position of M . p9 M83) Now take MN (40 p10mm) as fixed distance in M9compass, M1 center cut p11 M10line CB in N1. p124) Mark point P1 on M11M1N1 with same distance p13 M12of MP from M1.5) Similarly locate M13M2P2, M3P3, M4P4 and join D 52all P points. It
- 53. Problem No.7:A Link OA, 80 mm long oscillates around O, 60 0 to right side and returns toit’s initial vertical Position with uniform velocity.Mean while pointP initially OSCILLATING LINKon O starts sliding downwards and reaches end A with uniform velocity.Draw locus of point P p O Solution Steps:Point P- Reaches End A (Downwards) p11) Divide OA in EIGHT equal parts and 1 p2 p4 p3from O to A after O name 1, 2, 3, 4 up to 28. (i.e. up to point A).2) Divide 600 angle into four parts (150 3each) and mark each point by A1, A2, A3, p5 A4A4 and for return A5, A6, A7 andA8. 4(Initial A point). 5 p63) Take center O, distance in compass A3O-1 draw an arc upto OA1. Name this 6 A5point as P1. 7 p7 A21) Similarly O center O-2 distance A6mark P2 on line O-A2. 8 A A12) This way locate P3, P4, P5, P6, P7 and p8 A7P8 and join them. A8 ( It will be thw desired locus of P ) 53
- 54. Problem No 8: OSCILLATING LINK A Link OA, 80 mm long oscillates around O, 600 to right side, 1200 to left and returns to it’s initial vertical Position with uniform velocity.Mean while point P initially on O starts sliding downwards, reaches end A and returns to O again with uniform velocity.Draw locus of point PSolution Steps:( P reaches A i.e. movingdownwards. Op & returns to O again i.e.moves 16upwards ) 151.Here distance traveled by point P p1 p4 1 p2is PA.plus AP.Hence divide it into 14 p3eight equal parts.( so total linear 2 13displacement gets divided in 16parts) Name those as shown. A12 3 p5 12 A42.Link OA goes 600 to right, comes 4back to original (Vertical) position, 11goes 600 to left and returns to A11 5 p6 A13 A3original vertical position. Hence 10 A5total angular displacement is 2400. 6Divide this also in 16 parts. (150 A10 9 7 p7 A2each.) A14 A6Name as per previous problem.(A, A9 8 A1 A15 A p8 A7A1 A2 etc) A83.Mark different positions of P as A16per the procedure adopted in 54previous case.and complete the problem.
- 55. Problem 9: ROTATING LINKRod AB, 100 mm long, revolves in clockwise direction for one revolution.Meanwhile point P, initially on A starts moving towards B and reaches B.Draw locus of point P. A2 1) AB Rod revolves around center O for one revolution and point P slides along AB rod and A1 reaches end B in one A3 revolution. p1 2) Divide circle in 8 number of p2 p6 p7 equal parts and name in arrow direction after A-A1, A2, A3, up to A8. 3) Distance traveled by point P is AB mm. Divide this also into 8 p5 number of equal parts. p3 p8 4) Initially P is on end A. When A moves to A1, point P goes A B A4 P 1 4 5 6 7 one linear division (part) away 2 3 p4 from A1. Mark it from A1 and name the point P1. 5) When A moves to A2, P will be two parts away from A2 (Name it P2 ). Mark it as above from A2. 6) From A3 mark P3 three parts away from P3. 7) Similarly locate P4, P5, P6, A7 A5 P7 and P8 which will be eight parts away from A8. [Means P has reached B]. 8) Join all P points by smooth A6 55 curve. It will be locus of P
- 56. Rod AB, 100 mm long, revolves in clockwise direction for one revolution.Meanwhile point P, initially on A starts moving towards B, reaches B ROTATING LINK And returns to A in one revolution of rod. Draw locus of point P. Solution A21) AB Rod revolves around Stepscenter O for one revolution andpoint P slides along rod ABreaches end B and returns to A. A1 A32) Divide circle in 8 number ofequal parts and name in arrowdirection after A-A1, A2, A3, up toA8. p53) Distance traveled by point P p1is AB plus AB mm. Divide AB in 4parts so those will be 8 equalparts on return.4) Initially P is on end A. When p4A moves to A1, point P goes one p2 A A4linear division (part) away from P 1+7 2+6 p + 5 3 4 +BA1. Mark it from A1 and name the p8 6point P1.5) When A moves to A2, P willbe two parts away from A2 (Nameit P2 ). Mark it as above from A2.6) From A3 mark P3 three parts p7 p3away from P3.7) Similarly locate P4, P5, P6,P7 and P8 which will be eightparts away from A8. [Means P has A7 A5reached B].8) Join all P points by smoothcurve. It will be locus of P A6 56 TheLocus will follow the loop path twotimes in one revolution.

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