1. ENGINEERING CURVES Part-II (Point undergoing two types of displacements)INVOLUTECYCLOID SPIRALHELIX1. Involute of a circle 1. General Cycloid 1. Spiral of1. On Cylindera)String Length = DOne Convolution.2. Trochoid2. On a Cone b)String Length > D( superior) 2. Spiral of3. TrochoidTwo Convolutions. c)String Length < D( Inferior)4. Epi-Cycloid2. Pole having Composite shape.5. Hypo-Cycloid3. Rod Rolling overa Semicircular Pole. ANDMethods of DrawingTangents & Normals To These Curves.

2. DEFINITIONSCYCLOID: IS A LOCUS OF A POINT ON THE SUPERIORTROCHOID:ERIPHERY OF A CIRCLE WHICHIF THE POINT IN THE DEFINATIONOLLS ON A STRAIGHT LINE PATH. OF CYCLOID IS OUTSIDE THECIRCLENVOLUTE:INFERIOR TROCHOID.:IF IT IS INSIDE THE CIRCLEIS A LOCUS OF A FREE END OF A STRINGHEN IT IS WOUND ROUND A CIRCULAR POLEEPI-CYCLOIDIF THE CIRCLE IS ROLLING ONSPIRAL: ANOTHER CIRCLE FROM OUTSIDE IS A CURVE GENERATED BY A POINTHYPO-CYCLOID.HICH REVOLVES AROUND A FIXED POINTIF THE CIRCLE IS ROLLING FROMND AT THE SAME MOVES TOWARDS IT.INSIDE THE OTHER CIRCLE,HELIX: IS A CURVE GENERATED BY A POINT WHICHOVES AROUND THE SURFACE OF A RIGHT CIRCULARYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTIONT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION. or problems refer topic Development of surfaces) 3. Problem: Draw involute of an equilateral triangle of 35 mm sides.353X 5 2X3 35 3X3535 4. Problem: Draw involute of a square of 25 mm sides 75010 502525100 5. Problem no 17: 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 D distance away from A.Means if this string is wound roundthe circle, it will completely coverP2given circle. B will meet A afterwinding.2) Divide D (AP) distance into 8 P3number of equal parts.P13)Divide circle also into 8 number2 to pof equal parts. 34)Name after A, 1, 2, 3, 4, etc. up to pto 8 on D line AP as well as onpcircle (in anticlockwise direction). o 1t5)To radius C-1, C-2, C-3 up to C-8draw tangents (from 1,2,3,4,etc to4 to pcircle).P446)Take distance 1 to P in compass 3and mark it on tangent from point 15on circle (means one division less2than distance AP). 6 p o 5t7)Name this point P118)Take 2-P distance in compass7 A 8 6 to pand mark it on the tangent from7 P5toPpoint 2. Name it point P2.pP8 12 34 5 67 89)Similarly take 3 to P, 4 to P, 5 to P7P up to 7 to P distance in compassP6 Dand mark on respective tangentsand locate P3, P4, P5 up to P8 (i.e.A) points and join them in smoothcurve it is an INVOLUTE of a givencircle. 6. INVOLUTE OF A CIRCLEProblem 18: Draw Involute of a circle. String length MORE than DString length is MORE than the circumference of circle.Solution Steps: P2In this case string length is morethan D. But remember!Whatever may be the length of P3P1string, mark D distance2 to phorizontal i.e.along the stringand divide it in 8 number of3toequal parts, and not any other ppdistance. Rest all steps are sameo1tas previous INVOLUTE. Drawthe curve completely.4 to p P44352 po5t61 P5 7 87 p8 1 P 6 to pto p 2 34 5 67 8 D P7165 mm P6(more than D) 7. Problem 19: Draw Involute of a circle. INVOLUTE OF A CIRCLEString length is LESS than the circumference of circle.String length LESS than DSolution Steps:P2In this case string length is Lessthan D. But remember!Whatever may be the length ofP3P1string, mark D distancehorizontal i.e.along the stringand divide it in 8 number of2 to p 3 toequal parts, and not any otherpdistance. Rest all steps are sameas previous INVOLUTE. Drawop 1tthe curve completely.4 to p P44 3 5 2po65t 1 6 to pP57to7P p 8P7 1 2 34 56 78P6 150 mm (Less than D)D 8. PROBLEM 21 : Rod AB 85 mm long rollsover a semicircular pole without slippingfrom its initially vertical position till itbecomes up-side-down vertical.BDraw locus of both ends A & B. A4 Solution Steps?4If you have studied previous problems B1 properly, you can surely solve this also. Simply remember that this being a rod, A3it will roll over the surface of pole.3Means when one end is approaching, other end will move away from poll.OBSERVE ILLUSTRATION CAREFULLY!D2 A2B2 21 3 1 A14A B3 B4 9. Problem 22: Draw locus of a point on the periphery of a circle which rolls on straight line path .CYCLOIDTake circle diameter as 50 mm. Draw normal and tangent on the curve at a point 40 mmabove the directing line. 6 p5 p67 5p7 4 p4p8 8C1C2C3 C4 C5 C6C7 C8C9 C10 C11 p9 C129 p33 p2 p10 10 p12p11 11 1p1212 PDSolution Steps:1) From center C draw a horizontal line equal to D distance.2) Divide D distance into 12 number of equal parts and name them C1, C2, C3__ etc.3) Divide the circle also into 12 number of equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12.4) 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 up to C12 as centers. Mark points P2, P3, P4, P5 up to P12 on thehorizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively.7) Join all these points by curve. It is Cycloid. 10. PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A CURVED PATH. Takediameter of rolling Circle 50 mm 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 D distance on arc and it willbe decided by included arc angle .2)Calculate by formula = (r/R)x 3600.3)Construct angle with radiusOC and draw an arc by taking O as c8 c9c10center OC as radius and form c7 c11sector of angle . c12c64)Divide this sector into 12 numberof equal angular parts. And from C c5onward name them C1, C2, C3 up to89 10 11C12.7 6125)Divide smaller circle (Generating c4circle) also in 12 number of equal 5parts. And next to P in anticlockw-c3ise direction name those 1, 2, 3, up4to 12.6)With O as center, O-1 as radius c2 3draw an arc in the sector. Take O-2,O-3, O-4, O-5 up to O-12 distances 23with center O, draw all concentric42arcs in sector. Take fixed distancec11C-P in compass, C1 center, cut arc 5 1of 1 at P1. Repeat procedure and locate P 2, P3, 6C12P4, P5 unto P12 (as in cycloid) andPjoin them by smooth curve. This is O7 11 OP=Radius of directing circle=75mmEPI CYCLOID. PC=Radius of generating circle=25mm 810 =r/R X360= 25/75 X360=120 9 11. PROBLEM 26: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLEWHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter ofrolling 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 rotate9 anticlockwise to move7 8 1011 ahead.612 2)Same steps should be taken as in case of EPI 5 CYCLOID. Only change is 4 in numbering direction ofc7 c8 c9 c10c11c12 12 number of equal parts 3 c6 on the smaller circle.c5 3)From next to P in c4 clockwise direction, name 22 3 c34 1,2,3,4,5,6,7,8,9,10,11,12c2 4)Further all steps are 1 1 5 that of epi cycloid. Thisc1 is called HYPO CYCLOID. 12 C 6P 117O 10 8 9OP=Radius of directing circle=75mmPC=Radius of generating circle=25mm=r/R X360= 25/75 X360=120 12. Problem 27: Draw a spiral of one convolution. Take distance PO 40 mm. SPIRALIMPORTANT APPROACH FOR CONSTRUCTION!FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENTAND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2P2Solution Steps31P11. 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 OP 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 P5P6vector. 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. 6 13. Problem 28 SPIRALPoint P is 80 mm from point O. It starts moving towards O and reaches it in twoofrevolutions 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 P23,11P1 1,9 SOLUTION STEPS:P3 Total angular displacement here P10is two revolutions AndP9 Total Linear displacement hereP11 is distance PO. 1613108 7 6 5 4 3 2 1 P Just divide both in same parts i.e.4,12 P4P88,16P12 Circle in EIGHT parts.P15 ( means total angular displacementP13 P14in SIXTEEN parts) Divide PO also in SIXTEEN parts. P7 Rest steps are similar to the previousP5 problem. P6 5,137,15 6,14 14. 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 lmaMARK POINT OF INTERSECTION OF rNoTHIS SEMICIRCLE AND POLE CIRCLEAND JOIN IT TO Q. QTHIS WILL BE NORMAL TO INVOLUTE.TangDRAW A LINE AT RIGHT ANGLE TOenTHIS LINE FROM Q.tIT WILL BE TANGENT TO INVOLUTE.435C 2 6178 P P8 1 2 3 4 5 678D 15. 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.alNo r mIT WILL BE TANGENT TO CYCLOID. CYCLOID QTang e ntCP CC1C2 C3C4C5C6C7 C8P ND 16. Spiral. Method of Drawing Tangent & NormalSPIRAL (ONE CONVOLUSION.)2ntge No nTarm P2al3 1 Difference in length of any radius vectors Q P1 Constant of the Curve =Angle between the corresponding radius vector in radian. P3 OP OP2OP OP2 ==/2 1.574 P4 OP= 3.185 m.m. 7 6 5 4 3 2 1 P7 STEPS:*DRAW SPIRAL AS USUAL.P5P6 DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE CONSTANT OF CURVE CALCULATED ABOVE.* LOCATE POINT Q AS DISCRIBED IN PROBLEM AND5 7THROUGH IT DRAW A TANGENTTO THIS SMALLER CIRCLE.THIS IS A NORMAL TO THE SPIRAL.*DRAW A LINE AT RIGHT ANGLE6*TO THIS LINE FROM Q. IT WILL BE TANGENT TO CYCLOID. 17. 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 thatit always remains equidistant from point F and line AB. P7A P5SOLUTION STEPS:1.Locate center of line, perpendicular toP3 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 4F4 3 2 14.Take F-1 distance as radius and F ascenter draw an arccutting first parallel line to AB. Nameupper point P1 and lower point P2.P25.Similarly repeat this process by takingagain 5mm to right and left and locate P4P3 P4 .6.Join all these points in smooth curve.P6BP8 It will be the locus of P equidistance from line AB and fixed point F. 18. Problem 5:-Two points A and B are 100 mm apart. Basic Locus Cases: There is a point P, moving in a plane such that the difference of its distances from A and B always remains constant and equals to 40 mm. Draw locus of point P. p7p5 p3 p1Solution Steps:1.Locate A & B points 100 mm apart.2.Locate point P on AB line, P A B70 mm from A and 30 mm from B4 3 2 1 1 2 3 4As PA-PB=40 ( AB = 100 mm )3.On both sides of P mark points 5mm apart. Name those 1,2,3,4 as usual. p24.Now similar to steps of Problem 2,p4Draw different arcs taking A & B centersand A-1, B-1, A-2, B-2 etc as radius.p65. Mark various positions of p i.e. and joinp8 them in smooth possible curve.It will be locus of P 70 mm30 mm 19. Problem No.7:OSCILLATING LINKA Link OA, 80 mm long oscillates around O,600 to right side and returns to its initial verticalPosition with uniform velocity.Mean while pointP initially on O starts sliding downwards andreaches end A with uniform velocity.Draw locus of point PpO p1 Solution Steps: 1p2 p4 Point P- Reaches End A (Downwards)p3 1) Divide OA in EIGHT equal parts and from O to A after O 2 name 1, 2, 3, 4 up to 8. (i.e. up to point A). 2) Divide 600 angle into four parts (150 each) and mark each point by A1, A2, A3, A4 and for return A5, A6, A7 andA8. 3p5 A4 (Initial A point). 3) Take center O, distance in compass O-1 draw an arc upto4 OA1. Name this point as P1. 1) Similarly O center O-2 distance mark P2 on line O-A2.5p6 2) This way locate P3, P4, P5, P6, P7 and P8 and join them. A3 6A5( It will be thw desired locus of P ) 7 p7A2A6A8A1p8 A7A8 20. OSCILLATING LINK Problem No 8: A Link OA, 80 mm long oscillates around O, 600 to right side, 1200 to left and returns to its 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 P Op1615 p1 p4 1p2Solution Steps: 14p3( P reaches A i.e. moving downwards. 2 & returns to O again i.e.moves upwards ) 131.Here distance traveled by point P is PA.plus A 3 p5AP.Hence divide it into eight equal parts.( so 1212A4total linear displacement gets divided in 16 4parts) Name those as shown. 112.Link OA goes 600 to right, comes back to A 5 p6A13 11A3original (Vertical) position, goes 600 to left A510and returns to original vertical position. Hence 6total angular displacement is 2400. A10 p7A2Divide this also in 16 parts. (150 each.) 9 7 A14 A6Name as per previous problem.(A, A1 A2 etc)A9 8 A13.Mark different positions of P as per theA15 A p8procedure adopted in previous case. A7 A8and complete the problem. A16 21. ROTATING LINKProblem 9:Rod 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 andA1 reaches end B in oneA3 revolution. p1 2)Divide circle in 8 number ofp2p6p7 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.p3p8 4)Initially P is on end A. When A moves to A1, point P goesA B A4 P 1 4 56 7 one linear division (part) away2 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 curve. It will be locus of P 22. Problem 10 :ROTATING LINK Rod AB, 100 mm long, revolves in clockwise direction for one revolution. Meanwhile point P, initially on A starts moving towards B, reaches B And returns to A in one revolution of rod. Draw locus of point P.A2 Solution Steps1)AB Rod revolves around center OA1 A3for one revolution and point P slidesalong rod AB reaches end B andreturns to A.2)Divide circle in 8 number of equalp5parts and name in arrow direction p1after A-A1, A2, A3, up to A8.3)Distance traveled by point P is ABplus AB mm. Divide AB in 4 parts sothose will be 8 equal parts on return.p44)Initially P is on end A. When A p2A4Amoves to A1, point P goes one P1+72+6 p + 5 34 +Blinear division (part) away from A1.p86Mark it from A1 and name the pointP1.5)When A moves to A2, P will betwo parts away from A2 (Name itP2 ). Mark it as above from A2.p7 p36)From A3 mark P3 three partsaway from P3.7)Similarly locate P4, P5, P6, P7 A7and P8 which will be eight parts away A5from A8. [Means P has reached B].8)Join all P points by smooth curve.It will be locus of P The Locus will A6follow the loop path two times inone revolution.