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Point-path synthesis of antiparallelogram and kite four-bar linkages
Mechanism and Machine Theory, 1972, VOL 7, pp. 55-62.
Pergamon Press
Printed in Great Britain
Point-Path Synthesis of Antiparallelogram and Kite Four-Bar Linkages T. Koetsier* and B. Roth¢ Received I March 1971 Abstract When applied to point-path synthesis problems, at most seven design parameters are available for the dimensional synthesis of four-bar linkages with two sets of equal links in the four-bar chain. It is shown that there exist at most 36 such four-bars with coupler paths that pass through seven specified points. A method for completely determining all solutions is given. Zusammenfassung- Punktbahn-Synthese von Gelenksantiparallelogrammen und -deltoiden: T. Koetsier und B. Roth. Bei Viergelenksketten mit zwei Paaren gleich langer Glieder stehen for Probleme :ler Punktbahn-Synthese h6chstens sieben Entwurfsparameter zur Verf5gung. Es wird gezeigt, dab es h6ohstens 36 solcher Gelenkvierecke gibt, die Koppelkurven erzeugen, welche durch sieben gegebene Punkte gehen. Eine Methode zur Auffindung s~mtlicher L6sungen wird auseinandergesetzt. Pe~oMe--CHHTE3 AHTHFIAPAJ~EYIOFPAMMA H HETblPEX3BEHHHKOB THFIA ~YIETYH)) nO TOHKAMTPAEKTOPHI~.T. Kerc~ep, B. POT. ]lpH IlpRMeHCHRH Mffro~xa Tpacrropml~TO~ICKB BOHDOCaX CRHTe3a MOXHO HOJly~lTb He 60.rle~ CeMH ~OcTyrlHblX KOHCTpyKTIIBHblX napaMeTpos ~ a s reoMeTpHqecxoro CHHTe3a qeTblDCX3BeHHHKOB C ,,~ByM~ r"pymlaMH paBHblX 3BCHI~B B qe'1~lpeX3BeHHRKe, noKa3aHO, tlTO M a K C H M ~ I b H O ~ qH~"lO cyweCTByIOUIHX qeTIdpex3BeHHKKOB, B KOTOpblx T p a e x ' r o p ~ maTyHOs npoxoa ~ T qcpe3 7 onpe ~eJleHHl~X TOqeK, ~ 36. I'lpe~ox~eH MeTO~ ~'~H no~Horo onpe~eaex~Lq s c e x pem¢h'aR.
1. Introduction IT HAS been shown [1] that, in general, a planar four-bar linkage can be synthesized so that its coupler curve passes through nine arbitrary points. The synthesis problem is highly non-linear and, for each set of nine points, many solutions exist. Although it is rather easy to obtain several solutions[l], no method is known which yields all fourbars with coupler curves through a given set of nine points. In this paper we consider a special case of this general problem. Here we restrict ourselves to four-bar chains made up of two pairs of equal links. Such linkages can be synthesized so that a point in the c~upleroplane passes through (at most) seven arbitrary points. Because of the special nature of such four-bars a complete solution to the sevenopoint synthesis problem can be obtained. In this paper we prove that there may be at most thirty-six *Research Assistant.
~:Professor,Departmentof MechanicalEngineering,StanfordUniversity,Stanford,Califorhia,U.S.A.
55
56 such four-bars through seven specified points, and we show how to determine the linkage dimensions. Ruling out the parallelogram (since its coupler curves are always circles) there are two possible configurations: If the non-adjacent links are of equal length, and two links are always crossed, we have the so-called crossed parallelogram or antiparallelogram four-bar (Fig. la). If the equal links are adjacent we have the "kite" four-bar configuration* (Fig. lb). These two types of four-bar linkages are related by means of Roberts' theorem. In fact, an antiparallelogram four-bar has two kite four-bars as cognates. This is illustrated in Fig. 2 where ACC~A2 is an antiparallelogram and ABBIA1 and BCC~.B~. are Ai
81
(a)
Figure l a . Antiparalleiogram four-bar linkage. (AB = A1BI, ~ 1 = BB~).
(b)
Figure lb. Kite type four-bar linkage. (AA1 = AB, AtB1 = B~B).
~
2
Figure 2. An antiparallelogram four-bar and its two kite type cognates. *This linkage is also called an isosceles or rhomboid four-bar. It is s o m e t i m e s called a " s p e a r " if the kite
is in the folded-inposition.
57 kite mechanisms. Clearly, any coupler curve generated by an antiparallelogram may be generated in two ways by a kite four-bar. Since kite and antiparaUelogram four-bars are cognates, we can restrict ourselves to the antiparallelogram configuration. Once we find all antiparallelogram " mechanisms that possess a coupler curve through seven given points, application of Roberts' theorem gives us all the kite mechanisms with the same property.
2. The Antlparallelogram The coupler curves of an antiparallelogram mechanism are pedal curves* of ellipses or hyperbolas. This is well known and can be shown easily: (a) Assume an antiparallelogram mechanism with AB < AAI (Fig. la) and AB fixed. Because AB1AIB ~ AABA1 we have /_BB1A; = - / _ A ~ A B and this means, together with /_B]PA] = - / _ A P B and AB = AIB~, that AABP - AB~A~P. This last property yields A P = B ] P which means that B P + A P = B P + B ~ P = B B ] . Hence B P + A P is constant during the motion of the mechanism. Since P is the instantaneous rotation center we have the well-known result that the fixed centrode is an ellipse E with A and B as foci. Moreover it is obvious that the moving centrode is an ellipse E' in the moving plane with foci A~ and B~. The two ellipses are congruent and symmetric with respect to their common tangent in every position. This means that if we reflect the moving ellipse about the common tangent its image always coincides with the fixed ellipse. Similarly, an arbitrary point Q, in the moving plane reflected in every position about the common tangent, will always have its image at a fixed point Q0 in the fixed plane. Conversely this means that we can obtain the coupler curve of Q by reflecting Q0 (the basepoint of Q) about all the tangents of the fixed centrode E.t Co) In the case of an antiparallelogram with AB > AA1 and AB fixed, we have a similar situation; except in this case the fixed and moving centrodes are hyperbolas. The coupler curve of an arbitrary point Q in the moving plane can again be obtained by reflecting the basepoint Qo of Q in all the tangents to the fixed centrode. In this case the coupler curves are pedal curves of hyperbolas. In both cases (a) and Co) the foci A and B of the fixed centrode are the base-points of At and B~ respectively.
3. Approach to the Problem Let (pt,qe)i = 1,7 be the coordinates of the seven given points (2,, i : 1,7, with respect to some coordinate system in the fixed plane. Our task is to find all antiparallelogram mechanisms that have a coupler curve that goes through these seven points. Let the point Q0 = (x, y) be the base point of (2, i = 1,7, relative to a conic E. Then the seven perpendicular bisectors of QoQ, i = 1,7, are tangents to the conic E. From the discussion in the previous section it follows that E will be the fixed centrode of an antiparallelogram four-bar provided E is either an ellipse or a hyperbola. Also, once E is determined the linkage is uniquely defined. Hence, we can find a solution to the synthesis problem if we can determine a base point Qo. In what follows we consider Q0 as an unknown and seek to determine those Qo for which the seven perpendicular bisectors of Q~Qo, i = 1,7, are all tangent lines to the same *A pedal (or foot point) curve is defined as the locus of the projections of some point on all tangents to some curve. ~rlf we translate each tangent line, of E, parallel to itself until it passes through the corresponding position
of Q, the aggregateof all such lines will envelopean ellipse, Es (the distancefrom any point on E~ to Q0 is twice that of the correspondingpoint on E). Each tangentline of E2 then containsone position of Q which is the projectionof Q0on that tangent.Hence,the locusof Q is a pedalcurve of ellipseF_~.
58 ellipse or hyperbola. We solve this seven position problem by superimposing two six position problems.
4. The Condition for Six Lines to be Tangent to a Conic is That They Belong to a Curve of the Second Class It has been shown [2] that the condition for six points to be on a conic section is ( Xmi y ~j -- Xmj y ~, ) ( X,nk Y ~, -- xm, Y =k ) = (x,uY,o -- x,uY,u) (x.kY,u -- x,uY.k) ( xroYnk -- x.kYn~) ( xn, Y,a -- x,uY,u)
(I)
( x=j Ymk -- Xm, Y~j) (xmi Yml - - Xml Y,n~)
where the coordinates of the six points are (xp,yp), p = i ,j ,k ,l ,m,n, and xtj = x~-xj, yo = y~-y~, etc. We can obtain the condition that six lines with line coordinates ( u p , v , , w , ) , p = i , j , k , l , m , n , belong to a curve of the second class from (1) by substituting xp = ( u p / w j , ) , Yv = ( v d w , ) , P = i , j , k , l , m , n . We find (2)
( m j i ) ( rnlk ) ( n k j ) ( nli ) -- ( nji ) ( nlk ) ( m k j ) ( m l i ) = 0
with (mji)
=
U,,Vm U~Vj Ul ,etc. I'Vrn
Wj
Wi
5. The Analytical Part of the Solution Let the line coordinates of the seven perpendicular bisectors of QoQ~ be u~,vi,w~, i = 1,7. We have then Ui = Pi - - X
|
vi ---- qt - - Y
I i = 1,7
(3)
W~ = - - ½ ( piZ + qi z) + ½(xZ + y z)
The left-hand side of (3) should be p,u,oiv~,piwi, where p~ is an arbitrary constant. However since equation (2) is homogeneous in these constants we can, for simplicity, set p, = 1. So that
(ijk) =
p~ -- x
q~ -- y
pj-x Pk - x
qj-y qk - Y
--½(Pi2+q'2) +½(xZ+y2) I _½(pjZ + qfl) +½(x2 +y2) . _ ½ ( p k Z + qkZ) + ½ ( ~ + y Z )
From which it follows that
(ijk) =
p~ -- x Pj--Pi Pk -- Pi
qi -- Y qj--qi vk -- vi
--½(pi2+qiZ)+½(~+y2) - - ½ ( Pfl + qfi ) "q'-½ ( pi 2 + qi '~) -
-
I •
½( pk z + qk 2) + ½( pi z + qt z )
Apparently (ijk) = 0 is the circle through Qi, Qj and Qk. In order to find the basepoints Qo we are interested in, we apply condition (2) twice. Firstly for n , m , j , l , i , k = 1,2,3,4,5,6, arid secondly for n , m , j , l , i , k -- 1,2,3,4,5,7. We find two curves:
59 (235)(246)(136)(154) - ( 2 3 6 ) ( 2 5 4 ) ( 1 4 6 ) ( 1 3 5 ) = 0
(4)
(235)(247)(137)(154)- (237)(254)(147)(135) = 0
(5)
From the equations (4) and (5) we can draw several conclusions: The two curves are both of the eighth order. The two curves are both fourfold circular. Moreover, curve (4) has double points at Q1, Q2, Q~, Q4, Q5 and Qs, while curve (5) has double points at Q1, Q2, Q3, Q,, Qs and Qr. Two eighth order curves have 64 points of intersection. In our case 32 points coincide with the isotropic points because of the circularity of the curves. These points are not solutions of our problem. Moreover, the two curves have 20 points of intersection at QI, Q2, Qa, Q4 and Qs. These points do not correspond to solutions of our problem either. In this case (4) and (5) are only satisfied because one of the perpendicular bisectors of ~ , i = 1,7, is indeterminate and not because the seven perpendicular bisectors belong to one curve of the second class. The conclusion is that, in general, there are at most 6 4 - 3 2 - 2 0 = 12 real base points Q0. Hence there are at most 12 real antiparallelogram mechanisms with a coupler curve through seven given points. Moreover, since according to Roberts' theorem each antiparallelogram mechanism has two cognate kite mechanisms, we can see that there at most 24 kite mechanisms with a coupler curve through seven given points.
6. The Numerical Part of the Solution In the numerical example (Table 1) the points of intersection Qo of the two curves (4) and (5) are obtained by first plotting the two curves (Figs. 4a and 4b). From the plot we determined the approximate points of intersection. Then Newton's method was used to refine the approximations. In this way we found four real points of intersection. For each of the four Q0 we determined the antiparallelogram mechanism involved in the following way. We determined the conic section E that touches the perpendicular bisectors of QoQ~, i = 1,5. This conic section is the fixed centrode E. The foci of the fixed centrode are the points A and B (Fig. 3). Reflection of A and B in the common perpendicular of QoQ~ gave us respectively BI and A~ in position j (j = 1,7). In this way we computed the lengths of the links AB, A~B~, AA~, BBI, A~Q and B~Q in each of the seven positions of the mechanism we synthesized. The variations in the value of the length of AAI = BBI in the seven positions provided us with a
IIo°
Figure 3. Centrodes of the antiparallelogram four bar, and a point Q in the coupler plane and its base point Qo.
Table 1.
2
(-2.59,22-51)
3
(4.21,15.86)
I 35-399
2 35.400
3 35-400
i AAI=BB~
I 30-811
2 30-811
3 30.811
Qo = ( - 3-2075, - 21.575), A = ( - 18.033, 0.6519), B = ( 11.975,-6.2861 ) AB =A~B~ = 30.799, A ~ Q = 2 1 - 5 4 7 , B t Q = 2 6 . 7 1 7
Solution 4
i A A , = BB,
4
4 30-811
4 35.400
4 117-53
4 39.892
(9.42,15.21)
(-20.473,-4"0203), B=(-25.135,-39.09)
A B = A , B ~ = 3 5 . 3 7 8 , A~Q=64.264, BIQ=31"849
Qo=(-34-747,24.451),A=
Solution 3
Qo=(13.192,5.2724),A=(81.318,63.489), B=(-16.932,-2-3821) A B = A t B ~ = 118.29, A ~ Q = 3 1 " 0 8 1 , B j Q = 8 9 " 6 1 2 i I 2 3 A A , = BB, il7.53 117'53 117.53
Solution 2
Qo = ( 4 . 8 8 6 1 . 4 . 9 5 0 9 ) . A = ( - 1 7 - 3 0 4 , 0 - 1 1 4 7 8 ) , B = ( 1 7 - 1 5 9 , - 0 " ! 4 6 8 ) A B = A ~ B I = 34-464, A ~ Q = 13"289, B t Q - - 2 2 " 7 1 1 i I 2 3 A A ~ = BBt 39-892 39-892 39-892
Solution I
1
(-11-8,26.66)
i
Q,(p,,q,)
5
5 30"810
5 35-398
5 117"53
5 39-891
(32-91,- I 1.43)
6
6 30"811
6 35-390
6 117"54
6 39-893
(2-77,- 25.77)
7
7 30-811
7 35-384
7 I17.53
7 39"891
(-28.94,25-06)
0
0~
61
Rgure 4a. Locus of base points Qo for positions 1,2,3,4,5,6. The points marked with squares denote intersections with the curve in Fig. 4bo
Q,
Rgure 4b. Locus of base points Qo for positions 1,2,3,4,5,7. The points marked with squares denote, intersections with the curve in Fig. 4a. These four marked points are the Qo for positions ! = 1,7. (The curves do also intersect in points Qi, i = 1,2,3,4,5.)
62 check of our method. In Table 1 we can see that the variations are small for the four solutions listed. Assuming we have indeed found all the real intersections of the curves given in Fig. 4, we conclude that for the set of seven points Qi, i = 1,7, given in Table 1 there are twelve real solutions. T h e s e are the four antiparallelograms given in Table 1 and their eight cognate kite four-bars.
Acknowledgement-This research was supported in part by the National Science Foundation.
References [I] ROTH B. and FREUDENSTEIN F., Synthesis of path generating mechanisms by numerical methods, Trans. Am. Soc. Mech. Engrs. 8b'B,298-306 (1964). [2] FREUDENSTEIN F., BOTTEMA O. and KOETSIER T., Finite conic-section Burmester theory, J. Mechanisms 4, 359-373 (1969).
Pergamon Press
Printed in Great Britain
Point-Path Synthesis of Antiparallelogram and Kite Four-Bar Linkages T. Koetsier* and B. Roth¢ Received I March 1971 Abstract When applied to point-path synthesis problems, at most seven design parameters are available for the dimensional synthesis of four-bar linkages with two sets of equal links in the four-bar chain. It is shown that there exist at most 36 such four-bars with coupler paths that pass through seven specified points. A method for completely determining all solutions is given. Zusammenfassung- Punktbahn-Synthese von Gelenksantiparallelogrammen und -deltoiden: T. Koetsier und B. Roth. Bei Viergelenksketten mit zwei Paaren gleich langer Glieder stehen for Probleme :ler Punktbahn-Synthese h6chstens sieben Entwurfsparameter zur Verf5gung. Es wird gezeigt, dab es h6ohstens 36 solcher Gelenkvierecke gibt, die Koppelkurven erzeugen, welche durch sieben gegebene Punkte gehen. Eine Methode zur Auffindung s~mtlicher L6sungen wird auseinandergesetzt. Pe~oMe--CHHTE3 AHTHFIAPAJ~EYIOFPAMMA H HETblPEX3BEHHHKOB THFIA ~YIETYH)) nO TOHKAMTPAEKTOPHI~.T. Kerc~ep, B. POT. ]lpH IlpRMeHCHRH Mffro~xa Tpacrropml~TO~ICKB BOHDOCaX CRHTe3a MOXHO HOJly~lTb He 60.rle~ CeMH ~OcTyrlHblX KOHCTpyKTIIBHblX napaMeTpos ~ a s reoMeTpHqecxoro CHHTe3a qeTblDCX3BeHHHKOB C ,,~ByM~ r"pymlaMH paBHblX 3BCHI~B B qe'1~lpeX3BeHHRKe, noKa3aHO, tlTO M a K C H M ~ I b H O ~ qH~"lO cyweCTByIOUIHX qeTIdpex3BeHHKKOB, B KOTOpblx T p a e x ' r o p ~ maTyHOs npoxoa ~ T qcpe3 7 onpe ~eJleHHl~X TOqeK, ~ 36. I'lpe~ox~eH MeTO~ ~'~H no~Horo onpe~eaex~Lq s c e x pem¢h'aR.
1. Introduction IT HAS been shown [1] that, in general, a planar four-bar linkage can be synthesized so that its coupler curve passes through nine arbitrary points. The synthesis problem is highly non-linear and, for each set of nine points, many solutions exist. Although it is rather easy to obtain several solutions[l], no method is known which yields all fourbars with coupler curves through a given set of nine points. In this paper we consider a special case of this general problem. Here we restrict ourselves to four-bar chains made up of two pairs of equal links. Such linkages can be synthesized so that a point in the c~upleroplane passes through (at most) seven arbitrary points. Because of the special nature of such four-bars a complete solution to the sevenopoint synthesis problem can be obtained. In this paper we prove that there may be at most thirty-six *Research Assistant.
~:Professor,Departmentof MechanicalEngineering,StanfordUniversity,Stanford,Califorhia,U.S.A.
55
56 such four-bars through seven specified points, and we show how to determine the linkage dimensions. Ruling out the parallelogram (since its coupler curves are always circles) there are two possible configurations: If the non-adjacent links are of equal length, and two links are always crossed, we have the so-called crossed parallelogram or antiparallelogram four-bar (Fig. la). If the equal links are adjacent we have the "kite" four-bar configuration* (Fig. lb). These two types of four-bar linkages are related by means of Roberts' theorem. In fact, an antiparallelogram four-bar has two kite four-bars as cognates. This is illustrated in Fig. 2 where ACC~A2 is an antiparallelogram and ABBIA1 and BCC~.B~. are Ai
81
(a)
Figure l a . Antiparalleiogram four-bar linkage. (AB = A1BI, ~ 1 = BB~).
(b)
Figure lb. Kite type four-bar linkage. (AA1 = AB, AtB1 = B~B).
~
2
Figure 2. An antiparallelogram four-bar and its two kite type cognates. *This linkage is also called an isosceles or rhomboid four-bar. It is s o m e t i m e s called a " s p e a r " if the kite
is in the folded-inposition.
57 kite mechanisms. Clearly, any coupler curve generated by an antiparallelogram may be generated in two ways by a kite four-bar. Since kite and antiparaUelogram four-bars are cognates, we can restrict ourselves to the antiparallelogram configuration. Once we find all antiparallelogram " mechanisms that possess a coupler curve through seven given points, application of Roberts' theorem gives us all the kite mechanisms with the same property.
2. The Antlparallelogram The coupler curves of an antiparallelogram mechanism are pedal curves* of ellipses or hyperbolas. This is well known and can be shown easily: (a) Assume an antiparallelogram mechanism with AB < AAI (Fig. la) and AB fixed. Because AB1AIB ~ AABA1 we have /_BB1A; = - / _ A ~ A B and this means, together with /_B]PA] = - / _ A P B and AB = AIB~, that AABP - AB~A~P. This last property yields A P = B ] P which means that B P + A P = B P + B ~ P = B B ] . Hence B P + A P is constant during the motion of the mechanism. Since P is the instantaneous rotation center we have the well-known result that the fixed centrode is an ellipse E with A and B as foci. Moreover it is obvious that the moving centrode is an ellipse E' in the moving plane with foci A~ and B~. The two ellipses are congruent and symmetric with respect to their common tangent in every position. This means that if we reflect the moving ellipse about the common tangent its image always coincides with the fixed ellipse. Similarly, an arbitrary point Q, in the moving plane reflected in every position about the common tangent, will always have its image at a fixed point Q0 in the fixed plane. Conversely this means that we can obtain the coupler curve of Q by reflecting Q0 (the basepoint of Q) about all the tangents of the fixed centrode E.t Co) In the case of an antiparallelogram with AB > AA1 and AB fixed, we have a similar situation; except in this case the fixed and moving centrodes are hyperbolas. The coupler curve of an arbitrary point Q in the moving plane can again be obtained by reflecting the basepoint Qo of Q in all the tangents to the fixed centrode. In this case the coupler curves are pedal curves of hyperbolas. In both cases (a) and Co) the foci A and B of the fixed centrode are the base-points of At and B~ respectively.
3. Approach to the Problem Let (pt,qe)i = 1,7 be the coordinates of the seven given points (2,, i : 1,7, with respect to some coordinate system in the fixed plane. Our task is to find all antiparallelogram mechanisms that have a coupler curve that goes through these seven points. Let the point Q0 = (x, y) be the base point of (2, i = 1,7, relative to a conic E. Then the seven perpendicular bisectors of QoQ, i = 1,7, are tangents to the conic E. From the discussion in the previous section it follows that E will be the fixed centrode of an antiparallelogram four-bar provided E is either an ellipse or a hyperbola. Also, once E is determined the linkage is uniquely defined. Hence, we can find a solution to the synthesis problem if we can determine a base point Qo. In what follows we consider Q0 as an unknown and seek to determine those Qo for which the seven perpendicular bisectors of Q~Qo, i = 1,7, are all tangent lines to the same *A pedal (or foot point) curve is defined as the locus of the projections of some point on all tangents to some curve. ~rlf we translate each tangent line, of E, parallel to itself until it passes through the corresponding position
of Q, the aggregateof all such lines will envelopean ellipse, Es (the distancefrom any point on E~ to Q0 is twice that of the correspondingpoint on E). Each tangentline of E2 then containsone position of Q which is the projectionof Q0on that tangent.Hence,the locusof Q is a pedalcurve of ellipseF_~.
58 ellipse or hyperbola. We solve this seven position problem by superimposing two six position problems.
4. The Condition for Six Lines to be Tangent to a Conic is That They Belong to a Curve of the Second Class It has been shown [2] that the condition for six points to be on a conic section is ( Xmi y ~j -- Xmj y ~, ) ( X,nk Y ~, -- xm, Y =k ) = (x,uY,o -- x,uY,u) (x.kY,u -- x,uY.k) ( xroYnk -- x.kYn~) ( xn, Y,a -- x,uY,u)
(I)
( x=j Ymk -- Xm, Y~j) (xmi Yml - - Xml Y,n~)
where the coordinates of the six points are (xp,yp), p = i ,j ,k ,l ,m,n, and xtj = x~-xj, yo = y~-y~, etc. We can obtain the condition that six lines with line coordinates ( u p , v , , w , ) , p = i , j , k , l , m , n , belong to a curve of the second class from (1) by substituting xp = ( u p / w j , ) , Yv = ( v d w , ) , P = i , j , k , l , m , n . We find (2)
( m j i ) ( rnlk ) ( n k j ) ( nli ) -- ( nji ) ( nlk ) ( m k j ) ( m l i ) = 0
with (mji)
=
U,,Vm U~Vj Ul ,etc. I'Vrn
Wj
Wi
5. The Analytical Part of the Solution Let the line coordinates of the seven perpendicular bisectors of QoQ~ be u~,vi,w~, i = 1,7. We have then Ui = Pi - - X
|
vi ---- qt - - Y
I i = 1,7
(3)
W~ = - - ½ ( piZ + qi z) + ½(xZ + y z)
The left-hand side of (3) should be p,u,oiv~,piwi, where p~ is an arbitrary constant. However since equation (2) is homogeneous in these constants we can, for simplicity, set p, = 1. So that
(ijk) =
p~ -- x
q~ -- y
pj-x Pk - x
qj-y qk - Y
--½(Pi2+q'2) +½(xZ+y2) I _½(pjZ + qfl) +½(x2 +y2) . _ ½ ( p k Z + qkZ) + ½ ( ~ + y Z )
From which it follows that
(ijk) =
p~ -- x Pj--Pi Pk -- Pi
qi -- Y qj--qi vk -- vi
--½(pi2+qiZ)+½(~+y2) - - ½ ( Pfl + qfi ) "q'-½ ( pi 2 + qi '~) -
-
I •
½( pk z + qk 2) + ½( pi z + qt z )
Apparently (ijk) = 0 is the circle through Qi, Qj and Qk. In order to find the basepoints Qo we are interested in, we apply condition (2) twice. Firstly for n , m , j , l , i , k = 1,2,3,4,5,6, arid secondly for n , m , j , l , i , k -- 1,2,3,4,5,7. We find two curves:
59 (235)(246)(136)(154) - ( 2 3 6 ) ( 2 5 4 ) ( 1 4 6 ) ( 1 3 5 ) = 0
(4)
(235)(247)(137)(154)- (237)(254)(147)(135) = 0
(5)
From the equations (4) and (5) we can draw several conclusions: The two curves are both of the eighth order. The two curves are both fourfold circular. Moreover, curve (4) has double points at Q1, Q2, Q~, Q4, Q5 and Qs, while curve (5) has double points at Q1, Q2, Q3, Q,, Qs and Qr. Two eighth order curves have 64 points of intersection. In our case 32 points coincide with the isotropic points because of the circularity of the curves. These points are not solutions of our problem. Moreover, the two curves have 20 points of intersection at QI, Q2, Qa, Q4 and Qs. These points do not correspond to solutions of our problem either. In this case (4) and (5) are only satisfied because one of the perpendicular bisectors of ~ , i = 1,7, is indeterminate and not because the seven perpendicular bisectors belong to one curve of the second class. The conclusion is that, in general, there are at most 6 4 - 3 2 - 2 0 = 12 real base points Q0. Hence there are at most 12 real antiparallelogram mechanisms with a coupler curve through seven given points. Moreover, since according to Roberts' theorem each antiparallelogram mechanism has two cognate kite mechanisms, we can see that there at most 24 kite mechanisms with a coupler curve through seven given points.
6. The Numerical Part of the Solution In the numerical example (Table 1) the points of intersection Qo of the two curves (4) and (5) are obtained by first plotting the two curves (Figs. 4a and 4b). From the plot we determined the approximate points of intersection. Then Newton's method was used to refine the approximations. In this way we found four real points of intersection. For each of the four Q0 we determined the antiparallelogram mechanism involved in the following way. We determined the conic section E that touches the perpendicular bisectors of QoQ~, i = 1,5. This conic section is the fixed centrode E. The foci of the fixed centrode are the points A and B (Fig. 3). Reflection of A and B in the common perpendicular of QoQ~ gave us respectively BI and A~ in position j (j = 1,7). In this way we computed the lengths of the links AB, A~B~, AA~, BBI, A~Q and B~Q in each of the seven positions of the mechanism we synthesized. The variations in the value of the length of AAI = BBI in the seven positions provided us with a
IIo°
Figure 3. Centrodes of the antiparallelogram four bar, and a point Q in the coupler plane and its base point Qo.
Table 1.
2
(-2.59,22-51)
3
(4.21,15.86)
I 35-399
2 35.400
3 35-400
i AAI=BB~
I 30-811
2 30-811
3 30.811
Qo = ( - 3-2075, - 21.575), A = ( - 18.033, 0.6519), B = ( 11.975,-6.2861 ) AB =A~B~ = 30.799, A ~ Q = 2 1 - 5 4 7 , B t Q = 2 6 . 7 1 7
Solution 4
i A A , = BB,
4
4 30-811
4 35.400
4 117-53
4 39.892
(9.42,15.21)
(-20.473,-4"0203), B=(-25.135,-39.09)
A B = A , B ~ = 3 5 . 3 7 8 , A~Q=64.264, BIQ=31"849
Qo=(-34-747,24.451),A=
Solution 3
Qo=(13.192,5.2724),A=(81.318,63.489), B=(-16.932,-2-3821) A B = A t B ~ = 118.29, A ~ Q = 3 1 " 0 8 1 , B j Q = 8 9 " 6 1 2 i I 2 3 A A , = BB, il7.53 117'53 117.53
Solution 2
Qo = ( 4 . 8 8 6 1 . 4 . 9 5 0 9 ) . A = ( - 1 7 - 3 0 4 , 0 - 1 1 4 7 8 ) , B = ( 1 7 - 1 5 9 , - 0 " ! 4 6 8 ) A B = A ~ B I = 34-464, A ~ Q = 13"289, B t Q - - 2 2 " 7 1 1 i I 2 3 A A ~ = BBt 39-892 39-892 39-892
Solution I
1
(-11-8,26.66)
i
Q,(p,,q,)
5
5 30"810
5 35-398
5 117"53
5 39-891
(32-91,- I 1.43)
6
6 30"811
6 35-390
6 117"54
6 39-893
(2-77,- 25.77)
7
7 30-811
7 35-384
7 I17.53
7 39"891
(-28.94,25-06)
0
0~
61
Rgure 4a. Locus of base points Qo for positions 1,2,3,4,5,6. The points marked with squares denote intersections with the curve in Fig. 4bo
Q,
Rgure 4b. Locus of base points Qo for positions 1,2,3,4,5,7. The points marked with squares denote, intersections with the curve in Fig. 4a. These four marked points are the Qo for positions ! = 1,7. (The curves do also intersect in points Qi, i = 1,2,3,4,5.)
62 check of our method. In Table 1 we can see that the variations are small for the four solutions listed. Assuming we have indeed found all the real intersections of the curves given in Fig. 4, we conclude that for the set of seven points Qi, i = 1,7, given in Table 1 there are twelve real solutions. T h e s e are the four antiparallelograms given in Table 1 and their eight cognate kite four-bars.
Acknowledgement-This research was supported in part by the National Science Foundation.
References [I] ROTH B. and FREUDENSTEIN F., Synthesis of path generating mechanisms by numerical methods, Trans. Am. Soc. Mech. Engrs. 8b'B,298-306 (1964). [2] FREUDENSTEIN F., BOTTEMA O. and KOETSIER T., Finite conic-section Burmester theory, J. Mechanisms 4, 359-373 (1969).