Curves2
Upcoming SlideShare
Loading in...5
×
 

Curves2

on

  • 2,880 views

 

Statistics

Views

Total Views
2,880
Views on SlideShare
2,880
Embed Views
0

Actions

Likes
0
Downloads
90
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Curves2 Curves2 Presentation Transcript

  • 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: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 STRING IF IT IS INSIDE THE CIRCLEWHEN IT IS WOUND ROUND A CIRCLE OR POLYGON EPI-CYCLOIDSPIRAL: IF THE CIRCLE IS ROLLING ON ANOTHER CIRCLE FROMIT IS A CURVE GENERATED BY A POINT OUTSIDEWHICH REVOLVES AROUND A FIXED POINTAND AT THE SAME MOVES TOWARDS IT. HYPO-CYCLOID. IF THE CIRCLE IS ROLLING FROM INSIDE THE OTHERHELIX: CIRCLE,IT IS A CURVE GENERATED BY A POINT WHICHMOVES AROUND THE SURFACE OF A RIGHT CIRCULARCYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTIONAT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION.( for problems refer topic Development of surfaces)
  • Problem: Draw involute of a square of 25 mm sides 75 C 0 B 10 50 D 25 A 25 100
  • Problem: Draw involute of an equilateral triangle of 35 mm sides. X 35 3 B 5 2X3 C A 35 3X35 35
  • Problem no 23: Draw Involute of a circle of 40 mm diameter. INVOLUTE OF A CIRCLE Also draw normal and tangent to it at a point 100 mm from the centre of the circle.Solution Steps:1) Point or end P of string AP is P3exactly πD distance away from A. TangeMeans if this string is wound round P4 ntthe circle, it will completely covergiven circle. B will meet A after P2winding.2) Divide πD (AP) distance into 12number of equal parts.3) Divide circle also into 12 numberof equal parts. P5 al4) Name after A, 1, 2, 3, 4, etc. up Normto 12 on πD line AP as well as oncircle (in anticlockwise direction). P15) To radius C-1’, C-2’, C-3’ up toC-12’ draw tangents (from 1’,2’,3’,4’,etc to circle).6) Take distance 1 to P in compass P6and mark it on tangent from point 1’ 6’on circle (means one division less 7’ 5’than distance AP). 8’ 4’7) Name this point P18) Take 2-P distance in compass 9’ c 3’and mark it on the tangent from 10’ 2’ Ppoint 2’. Name it point P2.79) Similarly take 3 to P, 4 to P, 5 to 11’ 1’ 12’ PP up to 11 to P distance in compass A 1 2 3 4 5 6 7 8 9 10 11 12and mark on respective tangents P11 Pand locate P3, P4, P5 up to P12 (i.e. A) 8 πD P10points and join them in smooth P9curve it is an INVOLUTE of a givencircle.
  • 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 alMARK POINT OF INTERSECTION OF rm NoTHIS SEMICIRCLE AND POLE CIRCLEAND JOIN IT TO Q. QTHIS WILL BE NORMAL TO INVOLUTE. Ta n geDRAW A LINE AT RIGHT ANGLE TO ntTHIS LINE FROM Q.IT 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
  • 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 t p7 n ge an p8 8 4 p4 T p9 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 9 C p3 3 No 40mm p2 p10 rm CP al 10 2 p1 p11 11 1 p12 12 P 1’ 2’ 3’ 4’ 5’ 6’ 7’ 8’ 9’ 10’ 11’ 12’ Q πDSolution Steps:1) From center C draw a circle of 50mm dia. and from point P draw a horizontal line PQ equal to πD length.2) Divide the circle in 12 equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12.3) Also divide the straight line PQ into 12 number of equal parts and after P name them 1’,2’,3’__ etc.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 P1.6) 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.
  • 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 N o rmIT WILL BE TANGENT TO CYCLOID. CYCLOID Q Tang ent CP C C1 C2 C3 C4 C5 C6 C7 C8 P N πD
  • 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.Also draw normal andtangent on the curve at 110mm from the centre of directing circle.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 360º. c9 c10 c83) Construct angle θ with radius OC c7 c11and draw an arc by taking O as c12center OC as radius and form sector c6of angle θ.4) Divide this sector into 12 c5 8 9 10number of equal angular parts. And Rolling circle or 7 11from C onward name them C1, C2, generating c4 6 12C3 up to C12. circle 55) Divide smaller circle (Generating c3circle) also in 12 number of equal 4 Tangent Normalparts. And next to P in anticlockw- Directing circleise direction name those 1, 2, 3, up 3to 12. c26) With O as center, O-1 as radius 2 3’draw an arc in the sector. Take O-2, 4’ 2’O-3, O-4, O-5 up to O-12 distances c1 1with center O, draw all concentric 5’ 1’arcs in sector. Take fixed distanceC-P in compass, C1 center, cut arc θ 6’of 1 at P1. C P 12’Repeat procedure and locate P2, P3, OP4, P5 unto P12 (as in cycloid) and 7’ 11’ OP=Radius of directing circle=75mmjoin them by smooth curve. This is PC=Radius of generating circle=25mmEPI – CYCLOID. 8’ 10’ θ=r/R X360º= 25/75 X360º=120º 9’
  • Problem 17: A circle of 50 mm diameter rolls on another circle of 175 mm diameter and outside it.Draw the curve traced by a point P on its circumference for one complete revolution.Also draw normaland tangent on the curve at 125 mm from the centre of directing circle.Draw a horizontal line OP of 87.5 mm and draw an arcwith O as centre and PO as radiusDraw a horizontal line CP of 25 mm and draw a circlewith C as centre and CP as radius.θ=(OP/PC) X 360º = (25/87.5) X 360º = 102.8º ≈103ºDivide the rolling circle in 8 equal partsAlso divide the angle in 8 equal parts using angle bisectors Directing circle C O Rolling circle or P θ=103º generating circle
  • 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. Alsodraw normal and tangent on the curve at a point 40mm from the centre ofdirecting circle Directing circle Solution Steps: 1) Smaller circle is rolling here, inside the larger circle. It has to rotate 9 10 anticlockwise to move 7 8 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 c5 on the smaller circle. 3) From next to P in c4 2 clockwise direction, name 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 Rolling circle or 10’ 9’ 8’ generating circle OP=Radius of directing circle=75mm PC=Radius of generating circle=25mm θ=r/R X360º= 25/75 X360º=120º
  • Problem 28: A point P moves towards another point O, 75 mm from itand reaches it while moving around it once. Its movement towards Obeing uniform with its movement around it. Draw the curve traced out SPIRALby point P.Important approach for constructionFind total angular and total linear displacement and divide both in to same number of equalparts. Total linear movement 75 mm. Total angular movement 360º 2’With OP radius & O as center draw a circle and divide it in EIGHT P2 parts. Name those 1’,2’,3’,4’, etc. 3’ 1’ P1Similarly 8’ up to divided line PO also in EIGHT parts and name those P3 1,2,3, starting from P.Take O-1 distance from OP line and draw an arc up to O1’ radius 4’ P4 O P 7 6 5 4 3 2 1 vector. Name the point P1 P7Similarly mark points P2, P3, P4 up to P5 P6 P8And join those in a smooth curve. It is a SPIRAL of one convolution. 5’ 7’ 6’
  • Draw an Archemedian spiral of one convolution, greatest and least radii being 115mm and 15mm respectively. Draw a normal and tangent to the spiral at a point 65 mm from the pole.Important approach for construction!Find total angular and total linear displacement and divide both in to same number of equalparts. Angular displacement =360º, Linear displacement = 100mm 3’Solution Steps1. With PO & QO radii draw two 4’ 2’ P3 circles and divide them in P2 twelve equal parts. Name those P4 1’,2’,3’,4’, etc. up to 12’2 .Similarly divided line PQ also in 5’ P1 1’ twelve parts and name those P5 1,2,3,-- as shown. Tan3. Take O-1 distance from OP line N g and draw an arc up to O1’ radius en t al P6 Norm c Q P vector. Name the point P1 6’4. Similarly mark points P2, P3, P4 O 12 11 10 9 8 7 6 5 4 3 2 1 P11 12’ up to P12 P7 P10 And join those in a smooth curve. P8 P9 It is a SPIRAL of one convolution. 7’ 11’C=(Rmax-Rmin)/No. of convolutions in radians= (115-15)/3.14 X 2 =15.92 8’ 10’ 9’
  • Draw an Archemedian spiral of one and half convolution, greatest and least radii being 115mm and 15 mm respectively. Draw a normal and tangent to the spiral at a point 70 mm from the pole.Important approach for constructionFind total angular and total linear displacement and divide both in to same number of equalparts. Total Angular displacement 540º. Total Linear displacement 100 mm 3’15’ 1 Draw a 115 mm long line OP. 16’4’ 2’14’ P3 2 Mark Q at 15 mm from O P2 P4 3 with O as centre draw two circles with OP and OQ radius 4 Divide the circle in 12 equal divisions and 17’ 5’ P1 1’13’ mark the divisions as 1’,2’ and so on up to 18’ P 5 5 Divide the line PQ in 18 equal divis- P15 P14 ions as 1,2,3 and so on upto 18 P16 P13 6.Take O-1 distance from OP line and P17 draw an arc up to O1’ radius vector. P6 P12 Name the point P1 18’ 6’ Q P 7.Similarly mark points P2, P3, P4 up P18 O 18 16 14 12 10 8 6 4 2 12’ to P18. P11 8. And join those in a smooth curve. P7 It is a SPIRAL of one and half P10 convolution. P8 P9 7’ 11’C=(Rmax-Rmin)/No. of convolutions in radians= (115-15)/3.14 X3 =10.61 8’ 10’ 9’
  • Spiral. Method of Drawing Tangent & Normal SPIRAL (ONE CONVOLUSION.) 2 ent No ng Ta rm P2 a Difference in length of any radius vectors l 3 1 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.
  • 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
  • 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 1 p2 p4 p3 Solution Steps: 2 Point P- Reaches End A (Downwards) ) Divide OA in EIGHT equal parts and from O to A after O 3 p5 A4 name 1, 2, 3, 4 up to 8. (i.e. up to point A). 4 ) Divide 60 angle into four parts (15 each) and mark each 0 0 point by A1, A2, A3, A4 and for return A5, A6, A7 andA8. 5 p6 (Initial A point). A3 6 A5 ) Take center O, distance in compass O-1 draw an arc upto OA1. Name this point as P1. 7 p7 A2 ) Similarly O center O-2 distance mark P2 on line O-A2. A6 A8 A1 ) This way locate P3, P4, P5, P6, P7 and P8 and join them. p8 A7 A8 ( It will be thw desired locus of P )
  • 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 revolution. A3 2) Divide circle in 8 number of p1 equal parts and name in arrow p2 p6 p7 direction after A-A1, A2, A3, up to A8. 3) Distance traveled by point P is AB mm. Divide this also into 8 number of equal parts. p5 4) Initially P is on end A. When p3 p8 A moves to A1, point P goes one linear division (part) away A B A4 P 1 4 5 6 7 from A1. Mark it from A1 and 2 3 p4 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, P7 and P8 which will be eight A7 A5 parts away from A8. [Means P has reached B]. 8) Join all P points by smooth curve. It will be locus of P A6
  • 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 linear P 1+7 2+6 p + 5 3 4 +Bdivision (part) away from A1. Mark it p8 6from A1 and name the point P1.5) When A moves to A2, P will betwo parts away from A2 (Name itP2 ). Mark it as above from A2.6) From A3 mark P3 three parts p7 p3away from P3.7) Similarly locate P4, P5, P6, P7and P8 which will be eight parts away A7from A8. [Means P has reached B]. A58) Join all P points by smooth curve.It will be locus of P The Locus willfollow the loop path two times in A6one revolution.
  • Problem 28: A link OA, 100 mm long rotates about Oin anti-clockwise direction. A point P on the link, 15 θ= 2/5 X 360º = 144ºmm away from O, moves and reaches the end A, Total angular movement = 144ºwhile the link has rotated through 2/5 of a revolution. Total linear movement = 85 mmAssuming that the movements of the link to beuniform trace the path of point P. To divide both of them in equal no. of parts ( say 8) 5’ 6’ 4’ 7’ 3’ P6 P7 P5 8’ 2’ P8 P4 P3 1’ P2 144º P1 O P 1 2 3 4 5 6 7 8A 15 100
  • Logarithmic Spiral:If a point moves around a pole in such a way thatThe value of vectorial angle are in arithmatic progression andThe corresponding values of radius vectors are in geometric progression, then the curvetraced by the point is known as logarithmic spiral. A3 A2 P3 A1 P2 θ θ P1 θ O P A Let OA be a straight line and P be a point on it at radius vector OP from O. Now let the line moves at uniform angular speed to a new position OA1 ,at vectorial angle θ from OA and the point moves to a new position P1 , at radius vector OP1 from O. The line now gradually moves to the new position OA2, OA3 at vectorial angle θ and the point to P2 and P3 , at radius vectors OP2 and OP3 respectively. In Logarithmic spiral OP3/OP2 =OP2/OP1=OP1/OP
  • Problem37: In a logarithmic spiral, the shortest radius is 40mm. The length of adjacentradius vectors enclosing 30º are in the ratio of 9:8 Construct one revolution of the spiral.Draw tangent to the spiral at a point 70 mm from it.First step is to draw logarithmic scale. B P12Draw two straight lines OA & OB at angle of 30º. P11 P10Mark a point P on OA at 40 mm from O. P8 P9 P7Calculate OP1 such that OP1/OP = 9/8. => OP1 = 45 mm P P5 6 P4Mark a OP1 on OB at 45 mm from O. P2P3 P1Join P with P1. 45 P 4 P 3Draw an arc of radius OP1 from OB to OA. P2 O 30º A PDraw a line parallel to PP1 from P1 on OA to intersect OB at P2. 5 P P1P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P 40Repeat the steps to get the points P3,P4 and so on up to P12. 1 P6 P P12 P7 P11 P8 P9 P10