0
Upcoming SlideShare
×

# Curves2

186

Published on

Published in: Spiritual, Technology
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total Views
186
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
9
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Transcript of "Curves2"

1. 1. ENGINEERING CURVES Part-II (Point undergoing two types of displacements) INVOLUTE 1. Involute of a circle a)String Length = πD b)String Length > πD c)String Length < πD CYCLOID 1. General Cycloid 2. Trochoid ( superior) 3. Trochoid ( Inferior) 4. Epi-Cycloid SPIRAL HELIX 1. Spiral of One Convolution. 2. Spiral of Two Convolutions. 1. On Cylinder 2. On a Cone 2. Pole having Composite shape. 5. Hypo-Cycloid 3. Rod Rolling over a Semicircular Pole. AND Methods of Drawing Tangents & Normals To These Curves.
2. 2. DEFINITIONS CYCLOID: IS A LOCUS OF A POINT ON THE ERIPHERY OF A CIRCLE WHICH OLLS ON A STRAIGHT LINE PATH. NVOLUTE: SUPERIORTROCHOID: IF THE POINT IN THE DEFINATION OF CYCLOID IS OUTSIDE THE CIRCLE INFERIOR TROCHOID.: IS A LOCUS OF A FREE END OF A STRING IF IT IS INSIDE THE CIRCLE HEN IT IS WOUND ROUND A CIRCLE OR POLYGON SPIRAL: IS A CURVE GENERATED BY A POINT HICH REVOLVES AROUND A FIXED POINT ND AT THE SAME MOVES TOWARDS IT. EPI-CYCLOID IF THE CIRCLE IS ROLLING ON ANOTHER CIRCLE FROM OUTSIDE HYPO-CYCLOID. IF THE CIRCLE IS ROLLING FROM INSIDE THE OTHER CIRCLE, IS A CURVE GENERATED BY A POINT WHICH OVES AROUND THE SURFACE OF A RIGHT CIRCULAR YLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION T A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION. or problems refer topic Development of surfaces) HELIX:
3. 3. Problem: Draw involute of a square of 25 mm sides 75 C B 0 10 50 D 25 25 A 100
4. 4. Problem: Draw involute of an equilateral triangle of 35 mm sides. 35 3X B 5 2X3 C A 35 35 3X35
5. 5. Problem no 23: Draw Involute of a circle of 40 mm diameter. Also draw normal and tangent to it at a point 100 mm from the centre of the circle. P3 Ta n g e nt P4 al P2 Norm Solution Steps: 1) Point or end P of string AP is exactly πD distance away from A. Means if this string is wound round the circle, it will completely cover given circle. B will meet A after winding. 2) Divide πD (AP) distance into 12 number of equal parts. 3)  Divide circle also into 12 number of equal parts. P5 4)  Name after A, 1, 2, 3, 4, etc. up to 12 on πD line AP as well as on circle (in anticlockwise direction). 5)  To radius C-1’, C-2’, C-3’ up to C-12’ draw tangents (from 1’,2’,3’,4’, etc to circle). 6)  Take distance 1 to P in compass P6 and mark it on tangent from point 1’ on circle (means one division less than distance AP). 7)  Name this point P1 8)  Take 2-P distance in compass and mark it on the tangent from P point 2’. Name it point P2. 7 9)  Similarly take 3 to P, 4 to P, 5 to P up to 11 to P distance in compass and mark on respective tangents and locate P3, P4, P5 up to P12 (i.e. P8 A) points and join them in smooth curve it is an INVOLUTE of a given circle. INVOLUTE OF A CIRCLE 6’ 7’ 8’ 5’ 4’ 9’ c 3’ 10’ 2’ 11’ 12’ A P11 P9 P1 P10 1’ 1 P 2 3 4 5 6 πD 7 8 9 10 11 12
6. 6. Involute Method of Drawing Tangent & Normal STEPS: DRAW INVOLUTE AS USUAL. MARK POINT Q ON IT AS DIRECTED. INVOLUTE OF A CIRCLE ma l JOIN Q TO THE CENTER OF CIRCLE C. CONSIDERING CQ DIAMETER, DRAW A SEMICIRCLE AS SHOWN. No r MARK POINT OF INTERSECTION OF THIS SEMICIRCLE AND POLE CIRCLE AND JOIN IT TO Q. Q THIS WILL BE NORMAL TO INVOLUTE. Ta ng DRAW A LINE AT RIGHT ANGLE TO THIS LINE FROM Q. en t IT WILL BE TANGENT TO INVOLUTE. 4 3 5 2 C 6 7 1 8 P8 1 2 3 4 π D 5 6 7 P 8
7. 7. CYCLOID Problem 15: Draw locus of a point on the periphery of a circle which rolls on straight line path . Take circle diameter as 50 mm. Draw normal and tangent on the curve at a point 40 mm above the directing line. 7 5 a 4 p4 T 8 C1 p3 CP p2 10 11 p1 12 P 3 C3 2 1 1’ 2’ p7 p8 C4 C5 C6 al rm No C 9 C2 t en ng p6 p5 C7 C8 C9 C10 C11 p9 40mm 6 C12 p10 p11 p12 3’ 4’ 5’ 6’ πD 7’ 8’ 9’ 10’ 11’ 12’ Q Solution 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.
8. 8. STEPS: DRAW CYCLOID AS USUAL. MARK POINT Q ON IT AS DIRECTED. CYCLOID Method of Drawing Tangent & Normal WITH CP DISTANCE, FROM Q. CUT THE POINT ON LOCUS OF C AND JOIN IT TO Q. FROM THIS POINT DROP A PERPENDICULAR ON GROUND LINE AND NAME IT N JOIN N WITH Q.THIS WILL BE NORMAL TO CYCLOID. No r m al DRAW A LINE AT RIGHT ANGLE TO THIS LINE FROM Q. IT WILL BE TANGENT TO CYCLOID. CYCLOID Tang e nt CP Q C C1 C2 C3 C4 P C5 N πD C6 C7 C8
9. 9. PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A CURVED PATH. Take diameter of rolling Circle 50 mm And radius of directing circle i.e. curved path, 75 mm.Also draw normal and tangent on the curve at 110mm from the centre of directing circle. Tangent Solution Steps: 1)  When smaller circle will roll on larger circle for one revolution it will cover πD distance on arc and it will be decided by included arc angle θ. 2)  Calculate θ by formula θ = (r/R) x 360º. 3)  Construct angle θ with radius OC and draw an arc by taking O as center OC as radius and form sector of angle θ. 4)  Divide this sector into 12 number of equal angular parts. And from C Rolling circle or onward name them C1, C2, C3 up to generating circle C12. 5)  Divide smaller circle (Generating circle) also in 12 number of equal parts. And next to P in anticlockwise direction name those 1, 2, 3, up c2 to 12. 6)  With O as center, O-1 as radius 3’ 4’ draw an arc in the sector. Take O-2, O-3, O-4, O-5 up to O-12 distances c1 5’ with center O, draw all concentric arcs in sector. Take fixed distance C-P in compass, C1 center, cut arc 6’ C of 1 at P1. Repeat procedure and locate P 2, P3, 7’ P4, P5 unto P12 (as in cycloid) and join them by smooth curve. This is 8’ EPI – CYCLOID. 9’ c9 c8 c7 c10 c11 c12 c6 c5 8 9 10 7 12 6 c4 5 c3 Normal 4 3 Directing circle 2 2’ 1 1’ θ 12’ P 11’ 10’ 11 O OP=Radius of directing circle=75mm PC=Radius of generating circle=25mm θ=r/R X360º= 25/75 X360º=120º
10. 10. 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 normal and 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 arc with O as centre and PO as radius Draw a horizontal line CP of 25 mm and draw a circle with C as centre and CP as radius. θ=(OP/PC) X 360º = (25/87.5) X 360º = 102.8º ≈103º Divide the rolling circle in 8 equal parts Also divide the angle in 8 equal parts using angle bisectors Directing circle Rolling circle or generating circle C P O θ=103º
11. 11. PROBLEM 26: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter of rolling circle 50 mm and radius of directing circle (curved path) 75 mm. Also draw normal and tangent on the curve at a point 40mm from the centre of directing circle Directing circle Solution Steps: 1)  Smaller circle is rolling here, inside the larger circle. It has to rotate anticlockwise to move ahead. 2)  Same steps should be taken as in case of EPI – CYCLOID. Only change is in numbering direction of 12 number of equal parts on the smaller circle. 3)  From next to P in clockwise direction, name 1,2,3,4,5,6,7,8,9,10,11,12 4)  Further all steps are that of epi – cycloid. This is called HYPO – CYCLOID. 10 11 12 6 5 4 3 c5 c6 c7 c8 c9 c10 c11 c4 2 2’ 3’ c3 4’ c2 1 1’ 5’ c1 12’ θ 6’ C P 11’ Rolling circle or generating circle 9 8 7 7’ 10’ O 8’ 9’ OP=Radius of directing circle=75mm PC=Radius of generating circle=25mm θ=r/R X360º= 25/75 X360º=120º c12
12. 12. Problem 28: A point P moves towards another point O, 75 mm from it and reaches it while moving around it once. Its movement towards O being uniform with its movement around it. Draw the curve traced out by point P. SPIRAL Important approach for construction Find total angular and total linear displacement and divide both in to same number of equal parts. Total linear movement 75 mm. Total angular movement 360º With OP radius & O as center draw a circle and divide it in EIGHT parts. Name those 1’,2’,3’,4’, etc. up to divided line PO also in Similarly 8’ EIGHT parts and name those 1,2,3, starting from P. Take O-1 distance from OP line and draw an arc up to O1’ radius vector. Name the point P1 Similarly mark points P2, P3, P4 up to P8 And join those in a smooth curve. It is a SPIRAL of one convolution. 2’ P2 3’ P1 1’ P3 P4 4’ O P5 7 6 P7 5 4 3 2 1 P6 7’ 5’ 6’ P
13. 13. Draw an Archemedian spiral of one convolution, greatest and least radii being 115mm and 15 mm 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 equal parts. Angular displacement =360º, Linear displacement = 100mm 3’ Solution Steps 2’ 4’ P3 P2 P4 5’ P1 t gen 6’ P6 N al Norm c Q O 12 11 10 9 8 7 6 5 4 3 2 1 P11 P10 P7 P8 P9 7’ C=(Rmax-Rmin)/No. of convolutions in radians = (115-15)/3.14 X 2 =15.92 1’ P5 T an 1. With PO & QO radii draw two circles and divide them in twelve equal parts. Name those 1’,2’,3’,4’, etc. up to 12’ 2 .Similarly divided line PQ also in twelve parts and name those 1,2,3,-- as shown. 3. Take O-1 distance from OP line and draw an arc up to O1’ radius vector. Name the point P1 4. Similarly mark points P2, P3, P4 up to P12 And join those in a smooth curve. It is a SPIRAL of one convolution. 11’ 8’ 9’ 10’ P 12’
14. 14. 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 construction Find total angular and total linear displacement and divide both in to same number of equal parts. Total Angular displacement 540º. Total Linear displacement 100 mm 3’15’ 1 Draw a 115 mm long line OP. 16’4’ 2 Mark Q at 15 mm from O 3 with O as centre draw two circles with OP and OQ radius 4 Divide the circle in 12 equal divisions and 17’5’ mark the divisions as 1’,2’ and so on up to 18’ P 5 5 Divide the line PQ in 18 equal divisions as 1,2,3 and so on upto 18 6.Take O-1 distance from OP line and draw an arc up to O1’ radius vector. P6 Name the point P1 18’ 6’ 7.Similarly mark points P2, P3, P4 up to P18. P7 8. And join those in a smooth curve. It is a SPIRAL of one and half convolution. 7’ C=(Rmax-Rmin)/No. of convolutions in radians = (115-15)/3.14 X3 =10.61 8’ 2’14’ P3 P2 P4 P1 P16 P15 1’13’ P14 P13 P17 P18 P12 Q 18 O 16 14 12 10 8 6 4 2 P 12’ P11 P10 P8 P9 9’ 11’ 10’
15. 15. Spiral. Method of Drawing Tangent & Normal SPIRAL (ONE CONVOLUSION.) 2 No al rm 3 P2 nt ge n Ta P1 Q 1 Constant of the Curve = P3 = P4 4 O P5 7 6 P7 5 4 3 2 1 7 6 Angle between the corresponding radius vector in radian. OP – OP2 π/2 = OP – OP2 1.57 = 3.185 m.m. STEPS: *DRAW SPIRAL AS USUAL. DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE CONSTANT OF CURVE CALCULATED ABOVE. P6 5 P Difference in length of any radius vectors * LOCATE POINT Q AS DISCRIBED IN PROBLEM AND THROUGH IT DRAW A TANGENTTO THIS SMALLER CIRCLE.THIS IS A NORMAL TO THE SPIRAL. *DRAW A LINE AT RIGHT ANGLE *TO THIS LINE FROM Q. IT WILL BE TANGENT TO CYCLOID.
16. 16. SPIRAL of two convolutions Problem 28 Point P is 80 mm from point O. It starts moving towards O and reaches it in two revolutions around.it Draw locus of point P (To draw a Spiral of 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 P1 3,11 SOLUTION STEPS: Total angular displacement here is two revolutions And Total Linear displacement here is distance PO. Just divide both in same parts i.e. Circle in EIGHT parts. ( means total angular displacement in SIXTEEN parts) Divide PO also in SIXTEEN parts. Rest steps are similar to the previous problem. 1,9 P3 P10 P9 P11 4,12 P4 16 10 8 7 6 5 4 3 2 1 P P8 P12 8,16 P15 P13 P14 P7 P5 5,13 13 P6 6,14 7,15
17. 17. Problem No.7: A Link OA, 80 mm long oscillates around O, 600 to right side and returns to it’s initial vertical Position with uniform velocity.Mean while point P initially on O starts sliding downwards and reaches end A with uniform velocity. Draw locus of point P Solution Steps: Point P- Reaches End A (Downwards) 1) Divide OA in EIGHT equal parts and from O to A after O 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. (Initial A point). 3) Take center O, distance in compass O-1 draw an arc upto OA1. Name this point as P1. 1) Similarly O center O-2 distance mark P2 on line O-A2. 2) This way locate P3, P4, P5, P6, P7 and P8 and join them. ( It will be thw desired locus of P ) OSCILLATING LINK O p 1 2 p1 p2 p3 p4 3 p5 A4 4 5 p6 A3 6 7 8 A p8 A8 p7 A1 A7 A2 A6 A5
18. 18. 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 14 Solution Steps: ( P reaches A i.e. moving downwards. & returns to O again i.e.moves upwards ) 1.Here distance traveled by point P is PA.plus A 12 AP.Hence divide it into eight equal parts.( so total linear displacement gets divided in 16 parts) Name those as shown. 2.Link OA goes 600 to right, comes back to A A13 11 original (Vertical) position, goes 600 to left and returns to original vertical position. Hence total angular displacement is 2400. Divide this also in 16 parts. (150 each.) Name as per previous problem.(A, A1 A2 etc) 3.Mark different positions of P as per the procedure adopted in previous case. and complete the problem. 13 1 p1 p2 p3 p4 2 p5 3 12 A4 4 11 10 A10 A14 A3 6 9 7 A9 A15 p6 5 8 A p8 A8 A16 p7 A1 A7 A2 A6 A5
19. 19. ROTATING LINK Problem 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. 1)  AB Rod revolves around center O for one revolution and point P slides along AB rod and reaches end B in one revolution. 2)  Divide circle in 8 number of 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 number of equal parts. 4)  Initially P is on end A. When A moves to A1, point P goes one linear division (part) away 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, P7 and P8 which will be eight parts away from A8. [Means P has reached B]. 8)   Join all P points by smooth curve. It will be locus of P A2 A1 A3 p1 p2 p6 p5 A P 1 2 3 p7 p3 p4 A7 4 5 6 7 A5 A6 p8 B A4
20. 20. Problem 10 : 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. ROTATING LINK A2 Solution Steps 1)   AB Rod revolves around center O for one revolution and point P slides along rod AB reaches end B and returns to A. 2)   Divide circle in 8 number of equal parts and name in arrow direction after A-A1, A2, A3, up to A8. 3)   Distance traveled by point P is AB plus AB mm. Divide AB in 4 parts so those will be 8 equal parts on return. 4)   Initially P is on end A. When A moves to A1, point P goes one linear division (part) away 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, P7 and P8 which will be eight parts away from A8. [Means P has reached B]. 8)   Join all P points by smooth curve. It will be locus of P The Locus will follow the loop path two times in one revolution. A1 A3 p5 p1 p4 A P p8 p2 1+7 2+6 p 6 + 5 3 4 p7 p3 A7 A5 A6 +B A4
21. 21. Problem 28: A link OA, 100 mm long rotates about O in anti-clockwise direction. A point P on the link, 15 mm away from O, moves and reaches the end A, while the link has rotated through 2/5 of a revolution. Assuming that the movements of the link to be uniform trace the path of point P. θ= 2/5 X 360º = 144º Total angular movement = 144º Total linear movement = 85 mm To divide both of them in equal no. of parts ( say 8) 5’ 6’ 4’ 7’ 3’ P7 8’ P6 P5 2’ P4 P8 P3 1’ P2 144º P1 O 15 P 1 2 3 100 4 5 6 7 8A
22. 22. Logarithmic Spiral: If a point moves around a pole in such a way that The value of vectorial angle are in arithmatic progression and The corresponding values of radius vectors are in geometric progression, then the curve traced by the point is known as logarithmic spiral. A3 A2 P3 A1 P2 θ θ P1 θ O A P 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 OA 1 ,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
23. 23. Problem37: In a logarithmic spiral, the shortest radius is 40mm. The length of adjacent radius 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. Draw two straight lines OA & OB at angle of 30º. B Mark a point P on OA at 40 mm from O. Calculate OP1 such that OP1/OP = 9/8. => OP1 = 45 mm Mark a OP1 on OB at 45 mm from O. Join P with P1. P P Draw an arc of radius OP1 from OB to OA. P O P Draw a line parallel to PP1 from P1 on OA to intersect OB at P2. P Repeat the steps to get the points P3,P4 and so on up to P12. 4 45 P1 P P2 3 P4 P P5 6 P7 P8 P9 P10 P11 P12 3 2 30º 5 40 1 P6 P P12 P7 P11 P8 P9 P10 P P1P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 A
1. #### A particular slide catching your eye?

Clipping is a handy way to collect important slides you want to go back to later.