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.
DEFINITIONSCYCLOID: IS A LOCUS OF A POINT ON THE SUPERIORTROCHOID:ERIPHERY OF A CIRCLE WHICH IF THE POINT IN THE DEFINATIONOLLS ON A STRAIGHT LINE PATH. OF CYCLOID IS OUTSIDE THE CIRCLENVOLUTE: INFERIOR TROCHOID.: IF IT IS INSIDE THE CIRCLEIS A LOCUS OF A FREE END OF A STRINGHEN IT IS WOUND ROUND A CIRCULAR POLE EPI-CYCLOID IF THE CIRCLE IS ROLLING ONSPIRAL: ANOTHER CIRCLE FROM OUTSIDE IS A CURVE GENERATED BY A POINT HYPO-CYCLOID.HICH REVOLVES AROUND A FIXED POINT IF 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)
Problem: Draw involute of an equilateral triangle of 35 mm sides. 35 3X 5 2X3 35 3X35 35
Problem: Draw involute of a square of 25 mm sides 75 0 10 50 25 25 100
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 πD distance away from A.Means if this string is wound roundthe circle, it will completely covergiven circle. B will meet A afterwinding. P 3 2 to p2) Divide πD (AP) distance into 8 P2 1 Pnumber of equal parts.3) Divide circle also into 8 numberof equal parts.4) Name after A, 1, 2, 3, 4, etc. up 3 to pto 8 on πD line AP as well as oncircle (in anticlockwise direction). 1 to p5) 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 2 5 to pthan distance AP). 67) Name this point P1 18) Take 2-P distance in compass 7 A 8 7 to pand mark it on the tangent from Ppoint 2. Name it point P2. P8 1 2 3 4 5 6 7 8 5 P P6 6 to p9) Similarly take 3 to P, 4 to P, 5 to P 7P up to 7 to P distance in compass πDand 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.
INVOLUTE OF A CIRCLEProblem 18: Draw Involute of a circle. String length MORE than πDString length is MORE than the circumference of circle.Solution Steps:In this case string length is morethan Π D. P But remember! 2 to p P2 3Whatever may be the length of 1 Pstring, mark Π D distancehorizontal i.e.along the stringand divide it in 8 number ofequal parts, and not any other 3 to pdistance. Rest all steps are same 1 to pas previous INVOLUTE. Drawthe curve completely. 4 to p P4 4 3 5 5 to p 2 6 1 7 8 P 5 P p8 1 7 to p 2 3 4 5 6 7 8 πD P6 P 6 to p 7 165 mm (more than πD)
Problem 19: Draw Involute of a circle. INVOLUTE OF A CIRCLEString length is LESS than the circumference of circle. String length LESS than πDSolution Steps:In this case string length is Lessthan Π D. P 3 But remember! 2 to pWhatever may be the length of P2 1 Pstring, mark Π D distancehorizontal i.e.along the stringand divide it in 8 number ofequal parts, and not any other 3 to pdistance. Rest all steps are sameas previous INVOLUTE. Draw 1 to pthe curve completely. 4 to p P4 4 3 5 5 to p 2 6 1 7 7 to p 8 P P6 5 P 6 to p P 1 2 3 4 5 6 7 8 7 150 mm (Less than πD) πD
PROBLEM 21 : Rod AB 85 mm long rollsover a semicircular pole without slippingfrom it’s initially vertical position till itbecomes up-side-down vertical. BDraw locus of both ends A & B. A4 Solution Steps? 4 If you have studied previous problems B1 properly, you can surely solve this also. Simply remember that this being a rod, A3 it will roll over the surface of pole. 3 Means when one end is approaching, other end will move away from poll.OBSERVE ILLUSTRATION CAREFULLY! πD 2 A2 B2 2 1 3 1 A1 4 A B3 B4
Problem 22: Draw locus of a point on the periphery of a circle which rolls on straight line path . CYCLOID Take circle diameter as 50 mm. Draw normal and tangent on the curve at a point 40 mm above the directing line. 6 p5 p6 7 5 p7 4 p4 p8 8 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 p9 C12 9 p3 3 p2 p10 10 p1 2 p11 11 1 p12 12 P πDSolution Steps:1) From center C draw a horizontal line equal to πD distance.2) Divide πD 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 the horizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively.7) Join all these points by curve. It is Cycloid.
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 πD distance on arc and it willbe decided by included arc angle θ.2) Calculate θ by formula θ = (r/R)x 3600.3) Construct angle θ with radiusOC and draw an arc by taking O as c8 c9 c10center OC as radius and form c7 c11sector of angle θ. c12 c64) Divide this sector into 12 numberof equal angular parts. And from C c5onward name them C1, C2, C3 up to 8 9 10 11C12. 7 6 125) 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, up 4to 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 2 3’with center O, draw all concentric 4’ 2’arcs in sector. Take fixed distance c1 1C-P in compass, C1 center, cut arc 5’ 1’of 1 at P1. θRepeat procedure and locate P 2, P3, 6’ C 12’P4, P5 unto P12 (as in cycloid) and Pjoin them by smooth curve. This is O 7’ 11’ OP=Radius of directing circle=75mmEPI – CYCLOID. PC=Radius of generating circle=25mm 8’ 10’ θ=r/R X360º= 25/75 X360º=120º 9’
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 rotate 9 anticlockwise to move 7 8 10 11 ahead. 6 12 2) Same steps should be taken as in case of EPI – 5 CYCLOID. Only change is 4 in numbering direction of c7 c8 c9 c10 c11 c12 12 number of equal parts 3 c6 on the smaller circle. c5 3) From next to P in c4 clockwise direction, name 2 2’ 3’ c3 4’ 1,2,3,4,5,6,7,8,9,10,11,12 c2 4) Further all steps are 1 1’ 5’ that of epi – cycloid. This c1 is called θ HYPO – CYCLOID. 12’ C 6’ P 11’ 7’ O 10’ 8’ 9’ OP=Radius of directing circle=75mm PC=Radius of generating circle=25mm θ=r/R X360º= 25/75 X360º=120º
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. 6
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 6,14
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 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
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 πD
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.
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.
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 it’s distances from A and B always remains constant and equals to 40 mm. Draw locus of point P. p7 p5 p3 p1Solution Steps:1.Locate A & B points 100 mm apart.2.Locate point P on AB line, P A B 70 mm from A and 30 mm from B 4 3 2 1 1 2 3 4 As PA-PB=40 ( AB = 100 mm )3.On both sides of P mark points 5 mm apart. Name those 1,2,3,4 as usual. p24.Now similar to steps of Problem 2, p4 Draw different arcs taking A & B centers and A-1, B-1, A-2, B-2 etc as radius. p65. Mark various positions of p i.e. and join p8 them in smooth possible curve. It will be locus of P 70 mm 30 mm
Problem No.7: OSCILLATING LINKA Link OA, 80 mm long oscillates around O,600 to right side and returns to it’s initial verticalPosition with uniform velocity.Mean while pointP initially on O starts sliding downwards andreaches end A with uniform velocity.Draw locus of point P p O p1 Solution Steps: 1 p2 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. 3 p5 A4 (Initial A point). 3) Take center O, distance in compass O-1 draw an arc upto 4 OA1. Name this point as P1. 1) Similarly O center O-2 distance mark P2 on line O-A2. 5 p6 2) This way locate P3, P4, P5, P6, P7 and P8 and join them. A3 6 A5 ( It will be thw desired locus of P ) 7 p7 A2 A6 A8 A1 p8 A7 A8
OSCILLATING LINK Problem No 8: 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 P Op 16 15 p1 p4 1 p2Solution Steps: 14 p3( 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 12 12 A4total linear displacement gets divided in 16 4parts) Name those as shown. 112.Link OA goes 600 to right, comes back to A 5 p6 A13 11 A3original (Vertical) position, goes 600 to left A5 10and returns to original vertical position. Hence 6total angular displacement is 2400. A10 p7 A2Divide 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 the A15 A p8procedure adopted in previous case. A7 A8and complete the problem. A16
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 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 curve. It will be locus of P
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 O A1 A3for one revolution and point P slidesalong rod AB reaches end B andreturns to A.2) Divide circle in 8 number of equal p5parts 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 p2 A4 Amoves to A1, point P goes one P 1+7 2+6 p + 5 3 4 +Blinear division (part) away from A1. p8 6Mark 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.