A graphical method for solving the Euler-savary equation

Mech. Mach. Theory Vol. 25, No. 2, pp. 141-147, 1990 Printed in Great Britain. All rights reserved

0094-114X/90 $3.00 + 0.00 Copyright © 1990 Pergamon Press plc

A GRAPHICAL METHOD THE EULER-SAVARY GEORGE N. SANDOR, Y O N G X I A N

FOR SOLVING EQUATION X U and T Z U - C H E N WENG

Department of Mechanical Engineering, University of Florida, Gainesville, FL 32611, U.S.A.

(Received 21 August 1986; received for publication 15 September 1989)

Abstract--This paper presents a graphical method for solving the Euler-Savary equation (ESE) in all four of its forms. It is much simpler than Bobillier's construction for the solutions of the Euler-Savary equation in the fourth case.

INTRODUCTION

In the coplanar rolling of two conjugate curves on one another, the Euler-Savary equation (ESE) correlates the locations of instant center L an arbitrary point A of the plane of the moving curve, the instantaneous center O~ of the curvature of the path of this point described in the fixed plane, and the point on the ray of point A whose path momentarily has an infinite radius of curvature, the so-called inflection point, JA. Accordingly, the complex-number forms of the ESE, developed in Ref.[l], correlate the four points on a ray through the instant center, the point of tangency of the oscillating circles of the fixed and rolling centrodes, as shown in Fig. 1: 1. 2. 3. 4.

L the instant center. A, an arbitrary point of the moving rolling plane. JA, the inflection point on the ray A, also of the rolling plane. OA, the center of curvature of the instantaneous path of A in the vicinity of A, described in the fixed plane.

The four complex-number forms of the ESE provided the ways for digital computation to find one of the four points L A, JA and OA, when the other three are known. Graphically, Bobillier's construction can fulfill these task [2, 3]. However, it needs iteration to find point I when A, JA, and O~ are given. In this case, Bobillier's construction is not explicit and can not solve the ESE accurately. This paper presents a graphical method for the solutions of the ESE in all four cases. From the first of the complex-number forms of the ESE in Ref. [1]: a2

PA = Ia - JA [ exp(i arg(a -- Ja )),

ESE(I)

we can obtain the following two facts about the relationship of the four points on the ray, which are the basis of the method to be presented, i.e., (1) Points OA and Ja are always on the same side of A. (2) a 2 = (JAA)PA, for which a construction shown in Fig. 2 is needed, illustrating that h 2 =pq. THE

GRAPHICAL

METHOD

FOR

THE SOLUTION

OF T H E E S E

Case 1: Given I, A and OA on a Ray, Find JA (Fig. 3) Procedure 1. Draw line 1 through A normal to the ray. 2. Draw arc 2 through I centered at A to intersect line 1 at q. 3. Draw semicircle 3 through OA and q centered on the ray to intersect the ray at p. 141

GEORGEN. SANDORet al.

142

t

L_ FFig. 1. Rollcurve n rolls without slipping on the fixed curve P. Points 04, JA and A are on one ray from the instant center L

P " - - " ~

'~

q

-I

Fig. 2. Construction illustrating that h 2=pq.

q P

P

3

4

2 q

,4

PA

Fig. 3. Schematic for finding the inflection point J, when 1, A and 04 are given on the ray (Case 1).

Fig. 4. Schematic for finding O4 when L A and JA are given (Case 2).

4. D r a w semicircle 4 t h r o u g h p centered at A ; the intersection o f this semicircle with the ray, on the same side o f A as Oa, is the solution for JA.t

Proof F r o m the procedure o f the drawing, it is seen that a 2 = ~ ~ = (JA A)p~ a n d p o i n t s O4 a n d Ja are o n the same side o f AI which agrees with ESE(1). Therefore JA is the solution. Case 2: Given I, A and Ja on a Ray, Find OA (Fig. 4) Procedure 1. D r a w line 1 t h r o u g h A n o r m a l to the ray. 2. D r a w arc 2 t h r o u g h I centered at A to intersect line 1 at q. tlt is to be noted that this procedure for Case I was used on p. 74 in Ref. [4] for the solution of the inflection point, but there was no explanation why the inflection point was on that side of A shown in the procedure.

A graphical method forsoiving the ESE

143

3. Draw semicircle 3 through JA and q centered on the ray. The semicircle intersects the ray at p. 4. Draw semicircle 4 through p centered at A; the intersection of this semicircle with the ray, on the same side of A as JA, is the solution for Oa. P r o o f From the procedure, it is seen again that aS= q A ~ = (J~A)p~ and points OA and JA are on the same side of A, which agrees with ESE(1). Then point OA is the solution. Case 3: Given L OA and Ja on the Ray, Find A

(a) JA and

O. are on the same side of I (Fig. 5).

Procedure

1. Draw line 1 through I normal to the ray. 2. Draw semicircle 2 through JA centered at I to intersect the ray on the other side of I at t; thus ~ = n. 3. Draw semicircle 3 through t and O,. The semicircle intersects line 1 at q. 4. Draw arc 4 of radius m = IOA centered at t to intersect the ray such that the distance from the intersection p to I, i.e. P I = m + It = m + n, 5. Draw semicircle 5 through p and q centered on the ray; the intersection of this semicircle with the ray is the solution for A. P r o o f It is seen that from Step 3, ( ~ ) 2 = mn, and that from Step 5, ( ~ ) 2 = (m + n)a, where a = 1.4. Then we obtain

(2)

m n = (ra + n)a or

mn-

(m + n)a = 0

(3)

q

3

~"

\

~-

~

PA

(a) m > n q 2

4

Fig. 5. Schematic for finding A when I, Ya and OA are given. JA and Oa on the same side of I [Case 3(a)l.

4 A

(b) m < n

Fig. 6. Schematic for finding A when L JA and OA are given. JA and Oa on the opposite sides of I [Case 3(b)].

GEORGE N. SANDOR et al.

144

Adding a 2 to both sides of equation (3) yields mn-

(m + n )a + a 2 = a 2

or

a 2 = (n - a ) ( m - a).

Since m - a

= PA, n - - a = JAA, we have the condition a2 = (JA A)pA.

Moreover, from equation (2) a < m and a < n, which yields the fact that Ja and O4 are on the same side of A. These two facts agree with ESE(1). Therefore A is the solution of the ESE. (b) J~ and OA are on the opposite sides of I [Figs 6(a) and 6(b)]. Procedure

1. D r a w line 1 through I normal to the ray. 2. D r a w semicircle 2 through JA a n d OA. The semicircle 2 intersects line 1 at q. 3. Draw arc 3 of radius m = I 0 4 centered at JA to intersect the ray on the same side of JA as I, which makes the distance from the intersection p to I, i.e. Ip, equal to m - n when m > n [see Fig. 6(a)], or Ip = n - m when m < n [see Fig. 6(b)], where n=IJ

A.

4. Draw semicircle 4 through p and q centered on the ray; the intersection of this semicircle with the ray is the solution for A. Proof. It is seen that from Step 2, (~)~ = mn and that from Step 4, (-~)2 = (m - n)a when m > n [Fig. 6(a)], or (Iq) 2= (n - m ) a when m < n [Fig. 6(b)], where a = I-A. Then we have, when m>n,

mn=(m-n)a

(4)

or

a2+(m -n)a

-mn

= a2

or

a 2 = (a - n ) ( a + m).

Since a - n = J a A , and a + m = PA [Fig. 6(a)], we have a 2 = (JAA)pA. Moreover, it is seen that from equation (4), a > n, which yields the fact that JA and O~ are on the same side of A. These two facts agree with ESE(1), and therefore the point A is the solution of the ESE. In the other case, where m < n, we have (5)

m n = (n -- m ) a or

a 2+(n -m)a

-mn

= a2

or

a 2 = (a + n ) ( a - - m )

Since a + n = J A A , and a - m = PA [Fig. 6(b)], we obtain a 2 = (JAA)p~. Moreover, it is seen that from equation (5) a > m, which yields the fact that JA and OA are on the same side of A. These two facts agree with ESE(1). This completes the p r o o f that point A is the solution of the ESE. Case 4: Given Oa, A and JA on the Ray, Find I (Fig. 7) Procedure

1. Draw line 1 through A normal to the ray. 2. Draw semicircle 2 through JA centered at A, and m a r k its intersection with the ray, p. 3. D r a w semicircle 3 through p and OA. The semicircle intersects line 1 at q.

A graphical method for solving the ESE •

q

145

4

1)

Fig. 7. Schematic for finding double solutions of instant center I when A, JA and OA are given (Case 4).

4. Draw semicircle 4 centered at A, and its intersections with the ray on both sides of A are the two solutions of the ESE for instant center/, namely I ¢1) and p2).

Proof. From the procedure of the drawing, we know that for both positions of points I, namely Il and/2, a 2 = (~-~)2 = (J--~)PA and points Oa and JA are on the same side of A, which agrees with ESE(1). Then points I °) and I ¢2) are the two solutions for the instant center. EXAMPLE

Use the graphical method described in this paper to find the center of path curvature of coupler point C in the four-bar mechanism O~ABOs shown in Fig. 8.

dA

Fig. 8. Using the graphical method to find inflection points Ja and JB on the rays of A and B, respectively.

146

GEOROEN. SA~DORet al.

~

C

8

t I

n

Fig. 9. Drawingtwo lines through Ja and JB perpendicular to ray A and B, respectively,by whichthe inflectionpoleand the inflectioncircle are found.

Fig. 10. Using the graphical method to find 0 c when C, 1 and Jc are known.

In order to proceed with the solution, we will first find the instant center of the coupler with respect to the ground. Secondly, find the inflection points on the rays a and b, respectively. We can then draw the inflection circle, and find the inflection point on the ray of point C. Finally, we can obtain the center of the path curvature of point C. Procedure

1. Extend line AOA and line BOB; their intersection with each other is the instant center I of the coupler (Fig. 8). 2. Use the method described for Case 1 twice to find inflection points J~ and JB on the rays a and b, with known A, 04, I and B, OB, L respectively (Fig. 8). 3. Draw a line through JA normal to the ray a, and another line through Ja normal to the ray b. Then the intersection of these two lines is the inflection pole J, and the line IJ is the common normal of the centrodes (Fig. 9). 4. Draw the inflection circle of diameter IJ (Fig. 9). 5. Draw the ray c through C and L and extend it to intersect the inflection circle. The intersection Jc is the inflection point on the ray c (Fig. 9). 6. Now, points C, I and Jc are known on the ray c. Then using the graphical method described for Case 2, we finally find the center of the path curvature of coupler point C, namely 0 c (Fig. 10). CONCLUSION It is to be noted that, the above-described geometric constructions of the solutions to the ESE are of limited accuracy, especially if flat intersections render the graphical solution uncertain. Therefore, if exact results are required, computer solutions on the basis of the complex-number forms of the ESE, developed in Ref. [1], are recommended.

A graphical method for solving the ESE

147

Nevertheless, the readily visualizable results of the geometric procedures make this graphical approach a valuable tool. Acknowledgment--The authors wish to acknowledge the support of this research by the National Science Foundation at the University of Florida under Grant No. DMC-8505029. REFERENCES 1. G. N. Sandor, A. G. Erdman, L. Hunt and E. Raghavacharyulu, ASME J. mech. Des. 104, 227-232 (1982). 2. G. N. Sandor and A. G. Erdman, Advanced Mechanism Design--Analysis and Synthesis, pp. 320-327. Prentice Hall, NJ (1984). 3. G. N. Sandor, A. G. Erdrnan and E. Raghavacharyulu, Mech. Mach. Theory 20(2), 145-148 (1985). 4. R. Beyer, The Kinematic Synthesis of Mechanisms (English trans, by H. Kuenzel). McGraw-Hill, New York (1963).

GRAPHISCHE LOESUNG DER EULER--SAVARY GLEICHUNG Zusammenfassung--Dieser Auftrag behandlet eine graphische Methode fuer die Loesung der Euler-Savary Gleichung in alle vier ihrer Forrnen. Diese Methode ist viel einfacher als die Konstrukzion des Bobilliers fuer die Loesung der Euler-Savary Gleichung in den vierten Form.

M.M.T. 25/2--B