TThe Minkowskian planar 4R mechanism
G´abor Heged¨us
Johann Radon Institute for Computational and Applied Mathematics
Brian Moore
Advanced Telecommunications Research Institute International
April 17, 2019
Abstract
We characterize and classify completely the planar 4R closed chainworking on the Minkowskian plane. Our work would open a newresearch direction in the theory of geometric designs: the classificationand characterization of the geometric design of linkages working innon–Euclidean spaces.
First we recall here some preliminary definitions and results from the geom-etry of linkages.
A linkage is a collection of interconnected components, individually called links . The joint is the physical connection between two links.In our article we consider only the revolute joint, which can be viewed asconstructed from the rotary hinge. We denote the revolute joint by R.The revolute joint allows one-degree-of-freedom movement between thetwo links that it connects. The configuration variable for a hinge is the anglemeasured around its axis between the two bodies.Of course we can form linkages from other joints, for example the universaljoint, the ball-in-socket and the prismatic joint. Keywords. Mechanism, Minkowskian space, double numbers2010 Mathematics Subject Classification. 51P05, 53A17, 70B15 a r X i v : . [ m a t h . M G ] O c t he generic mobility of the system is the number of independent param-eters such as the joint angles that are needed to specify the configuration ofthe linkage.On the other hand it can be shown that this is the dimension of the configuration space of the system.A planar linkage has the property that all of its links move in parallelplanes. We are interested here in the four–bar linkage, which is a closedchain formed by four links and four joints. Figure 1 is an example of aplanar 4R closed chain. • •• • O CA B (cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23) (cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39) (cid:112) (cid:112) (cid:115) (cid:115) θ ψga h b
Figure 1: The planar 4R linkageThe usual model of geometric designs are working in the Euclidean space.In this article we would like to characterize and classify completely the planar4R closed chain working on the Minkowskian plane. This work would open anew research direction in the theory of geometric designs: the classificationand characterization of the geometric design of linkages working in non–Euclidean spaces.In Chapter 2 we collected the preliminary definitions and results aboutthe Minkowskian plane and hyperbolic trigonometry. In Chapter 3 we de-scribe our main results: the position analysis and the classification of theMinkowskian 4R planar linkage. We give also an exact formula for the cou-pler curve of this system and compute the transmission and coupler angles.2
Preliminaries
First we give an algebraic description of the Minkowskian plane.In analogy with the complex number system, the system of double num-bers can be introduced: H := { x + jy : x, y ∈ R , j = 1 } Here j is the double imaginary unit and x and y are respectively calledthe real and the unipotent parts of the double number z = x + yj .It follows that multiplication in H is defined by ( x + yj )( r + sj ) = ( xr + ys ) + j ( xs + yr ).It is known that the complex numbers are related to the Euclidean ge-ometry. Similarly the double system of numbers serve as coordinates in theMinkowskian plane (space-time geometry, see [5]).The hyperbolic conjugate z of z = x + yj is defined by z = x − yj .The hyperbolic scalar product is given by (cid:104) z, w (cid:105) := Re ( zw ) = xu − yv, where z = x + yj and w = u + jv . We say that the double numbers z and w are double–orthogonal , if (cid:104) z, w (cid:105) = 0.We define the hyperbolic modulus of z = x + yj by (cid:107) z (cid:107) h := (cid:112) |(cid:104) z, z (cid:105)| = (cid:112) | zz | = (cid:112) | x − y | ≥ . (1)We can consider this modulus as the hyperbolic distance of the point z from the origin.It can be shown that this modulus is the Lorentz invariant of two dimen-sional special relativity, see [12].Note that the points z (cid:54) = 0 on the lines y = x and y = − x are isotropic.This means that they are nonzero vectors with (cid:107) z (cid:107) h = 0.For r ∈ R + the Minkowskian circle of radius r centered at the origin in H is defined by { ( x, y ) ∈ R : x − y = r } Clearly this set is the set of all points in the Minkowskian plane that satisfythe equation (cid:107) z (cid:107) h = r (see Figure 2). these numbers are called to split–complex numbers, too Y • (cid:65) (cid:65) (cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2) (cid:47) (cid:47) (cid:79) (cid:79) x − y = r C ( r ch( t ) , r sh( t ))Figure 2: The Minkowskian circle of radius r centered at the origin4e can remind here for the following analog of Euler’s formula for thedouble numbers z = x + yj : z = re jφ = r (ch ( φ ) + j sh ( φ ))where x − y > y > Let L := L denote the vector space R provided with the hyperbolic scalarproduct.We denote by SO + (1 ,
1) the proper Lorentzian group consisting of allmatrices of the form A ( φ ) = (cid:18) ch ( φ ) sh ( φ )sh ( φ ) ch ( φ ) (cid:19) , where φ ∈ R , see [1], [12].In L a Lorentzian vector u is called to spacelike , lightlike or timelike if (cid:104) u, u (cid:105) L > (cid:104) u, u (cid:105) L = 0 or (cid:104) u, u (cid:105) L <
0, respectively.We say that a timelike vector u = ( u , u ) is future–pointing or past–pointing if u > u <
0, respectively. Similarly, a spacelike vector u =( u , u ) is future–pointing or past–pointing if u > u <
0, respectively.The following Lemma was proven in [1].
Lemma 2.1 (Reversed triangle inequality) Let x and y be future–pointingtimelike vectors in L . Then x + y is a future–pointing timelike vector and (cid:107) x + y (cid:107)≥(cid:107) x (cid:107) + (cid:107) y (cid:107) . Here the equality holds iff y = cx for some c > . Lemma 2.2
Let x and y be future–pointing spacelike vectors in L . Then x + y is a future–pointing spacelike vector and (cid:107) x + y (cid:107)≥(cid:107) x (cid:107) + (cid:107) y (cid:107) . Here the equality holds iff y = cx for some c > . roof. Suppose that x = −→ AB , y = −−→ BC and x + y = −→ AC . Now after reflectingthe points A , B and C to the line y = x we get the points A (cid:48) , B (cid:48) and C (cid:48) .Let d := −−→ A (cid:48) B (cid:48) and e := −−→ B (cid:48) C (cid:48) . Then d + e = −−→ A (cid:48) C (cid:48) and it comes from thedefinition of the norm that (cid:107) x (cid:107) = (cid:107) d (cid:107) , (cid:107) y (cid:107) = (cid:107) e (cid:107) and (cid:107) d + e (cid:107) = (cid:107) x + y (cid:107) . Hence using Lemma 2.1 we get that (cid:107) d + e (cid:107) = (cid:107) x + y (cid:107)≥(cid:107) x (cid:107) + (cid:107) y (cid:107) = (cid:107) d (cid:107) + (cid:107) e (cid:107) . Corollary 2.3
Suppose that
OABC is a quadrilaterial where a = (cid:107) −→ OA (cid:107) , b = (cid:107) −−→ BC (cid:107) , g = (cid:107) −→ OC (cid:107) , h = (cid:107) −→ AB (cid:107) . Suppose that −→ OA , −−→ BC , −→ OC and −→ AB are future–pointing spacelike vectors (see Figure 1). Then g ≥ a + b + h .Moreover, if g = a + b + h , then the vectors −→ OA , −−→ BC , −→ OC and −→ AB arecollinear. Proof.
The inequality g ≥ a + b + h is a direct consequence of Lemma 2.2.Now, assume that g = a + b + h . We prove first that the vectors −→ OC and −−→ OB are collinear. Suppose, indirectly, that −→ OC and −−→ OB are not collinear.Since −−→ OB is a future–pointing spacelike vector, hence we can apply the re-versed triangle inequality (Lemma 2.2) for the triangle OBC . Then it followsthat g = (cid:107) −→ OC (cid:107) > (cid:107) −−→ OB (cid:107) + (cid:107) −−→ BC (cid:107) = b + (cid:107) −−→ OB (cid:107) We can apply again Lemma 2.2 for the triangle
OAB . Hence (cid:107) −−→ OB (cid:107)≥ a + h. Consequently g = (cid:107) −→ OC (cid:107) > a + b + h, which is a contradiction.Similar argument shows that −−→ OB and −→ OA are colllinear. Hence −→ OC and −→ OA are collinear. We can use very similar arguments to prove that −→ OA and −→ AB are collinear vectors. So the result follows.6e define now the notion of angle on the Minkowskian plane.Let x and y be two future–pointing timelike unit vectors. We say that φ ∈ R is the (oriented) angle from x to y if A ( φ ) x = y . The (unoriented)angle between x and y is defined to be | φ | . Then it comes from the definitionthat ch ( φ ) = −(cid:104) x, y (cid:105) L , where the right-hand side is greater than 1.When x and y are future–pointing timelike vectors, than the angle φ for x and y is the same as for x/ (cid:107) x (cid:107) and y/ (cid:107) y (cid:107) .We have ch ( φ ) = − (cid:104) x, y (cid:105) L (cid:107) x (cid:107)(cid:107) y (cid:107) We can give the same definition for the angle between future–pointingspacelike vectors. From this definition, we obtain similar formula for ch ( φ ):if x and y are two future–pointing spacelike unit vectors, thench ( φ ) = (cid:104) x, y (cid:105) L , where φ is the (oriented) angle from x to y .In the Euclidean plane, a motion can be represented by a combination of arotation and translation. It is well–known that any motion can be expressedusing the matrix operation M ( t, a, b ) x x = cos( t ) − sin( t ) a sin( t ) cos( t ) b x x Similarly, for the Minkowskian plane the group of motions is the following: G := ch ( t ) sh ( t ) a sh ( t ) ch ( t ) b : t, a, b ∈ R We now introduce an important class of triangles. By a pure triangle wemean a triangle with vertices
A, B and C such that −→ AB and −−→ BC are future–pointing timelike vectors. In the following we assume that we named thevertices of a pure triangle ABC in this manner. The angle (cid:98) C is the anglebetween the lines BC and AC .Finally we recall here for the Minkowskian cosine rule:7 heorem 2.4 (see [1, Theorem 7]) Let (cid:52) ABC be a pure triangle. Then c = a + b − ab ch( (cid:98) C ) (2) where a = (cid:107) −−→ BC (cid:107) , b = (cid:107) −→ AC (cid:107) and c = (cid:107) −→ AB (cid:107) . Recall that the four-bar linkage is a mechanism that lies in the plane andconsists of four bars connected by joints that allow rotation only in the planeof the mechanism, see Figure 1.Throughout this Chapter we suppose that −→ OA , −→ OC and −−→ CB are future–pointing spacelike vectors. Suppose that −→ AB is a spacelike vector.Let the fixed and the moving pivots of the input crank be O and A ,respectively.Let the fixed and the moving pivots of the output crank be C and B ,respectively. We define the distances between these points as follows: a := (cid:107) −→ OA (cid:107) , b := (cid:107) −−→ BC (cid:107) , g := (cid:107) −→ OC (cid:107) , h := (cid:107) −→ AB (cid:107) . To analyse the linkage, we locate the origin in the fixed Minkowskianframe F at O and orient it so that the x -axis passes through the other fixedpivot C . Theorem 3.1
Let θ be the input angle measured around O from the x -axisof F to OA . Let ψ be the angular position of the output crank CB (see Figure1). Then ψ = 2 artanh − B ( θ ) ± (cid:112) B ( θ ) + C ( θ ) − A ( θ ) A ( θ ) + C ( θ ) , (3) where A ( θ ) = 2 gb − ab ch ( θ ) ,B ( θ ) = 2 ab sh ( θ ) , nd C ( θ ) = h − g − b − a + 2 ag ch ( θ ) . Proof.
Since h = (cid:107) −→ AB (cid:107) is constant, we get the constraint equation as (cid:107) −→ AB (cid:107) = h . (4)It is easy to verify that the coordinates of A and B is given by A =( a ch ( θ ) , a sh ( θ )) and B = ( g + b ch ( ψ ) , b sh ( ψ )) . (5)If we substitute these coordinates into (4), then we obtain b + g + a + 2 gb ch ( ψ ) − ag ch ( θ ) − ab ch ( ψ ) ch ( θ ) + 2 ab sh ( ψ ) sh ( θ ) = h . If we gather the coefficients of ch ( ψ ) and sh ( ψ ), we obtain the constraintequation for the 4R chain as A ( θ )ch ( ψ ) + B ( θ )sh ( ψ ) = C ( θ ) , (6)where A ( θ ) = 2 gb − ab ch ( θ ) ,B ( θ ) = 2 ab sh ( θ ) , and C ( θ ) = h − g − b − a + 2 ag ch ( θ ) . We solve this equation using the tan-half-technique. This technique usesa transformation of variables to convert ch ( ψ ) and sh ( ψ ) into algebraic func-tions of th ( ψ/ y = th ( ψ/ ψ ) = 1 + y − y and sh ( ψ ) = 2 y − y Substitute into (6) to obtain( A ( θ ) + C ( θ )) y + 2 B ( θ ) y + A ( θ ) − C ( θ ) = 0 . ψ/
2) = − B ( θ ) ± (cid:112) B ( θ ) + C ( θ ) − A ( θ ) A ( θ ) + C ( θ ) . Remark.
It can be derived the following alternative equation for ψ : ψ = − artanh B ( θ ) A ( θ ) ± arch C ( θ ) (cid:112) A ( θ ) − B ( θ ) . Namely we infer from (6) that A ( θ ) (cid:112) A ( θ ) − B ( θ ) ch ( ψ ) + B ( θ ) (cid:112) A ( θ ) − B ( θ ) sh ( ψ ) = C ( θ ) (cid:112) A ( θ ) − B ( θ ) . (7)Since A ( θ ) (cid:112) A ( θ ) − B ( θ ) ≥ , hence there exists δ such thatch ( δ ) = A ( θ ) (cid:112) A ( θ ) − B ( θ ) and sh ( δ ) = B ( θ ) (cid:112) A ( θ ) − B ( θ ) . Then clearly th ( δ ) = B ( θ ) A ( θ ) . We infer from (7) thatch( δ + ψ ) = ch ( δ ) ch ( ψ ) + sh ( δ ) sh ( ψ ) = C ( θ ) (cid:112) A ( θ ) − B ( θ ) . The result follows.
Remark.
A graph of ψ as a function of θ is called the (kinematic) trans-mission function . The transmission function has a very characteristic shape,10ence the type of the Minkowskian four-bar linkage can immediately be readfrom it. Remark.
Notice that there are two angles ψ for each angle θ . This arisesbecause the moving pivot B of the output crank can be assembled above orbelow the diagonal joining the moving pivot A of the input crank to the fixedpivot of C . The transmission function gives for any θ two, one, no or an infinite number ofvalues of ψ . No values for ψ exist iff the position of the linkage is impossible.In an ’undetermined position’ θ and ψ are independent of each other: at asingle θ , an infinite amount of values of ψ is possible or vice versa. We callthese θ angles to the branching points of the mechanism. Corollary 3.2 (Branching points) Suppose that g (cid:54) = b . Then the branchingpoints of the Minkowskian planar 4R occours at θ = arch (cid:18) a − h a ( g − b ) + g − b a (cid:19) . (8) Proof.
It is clear from Theorem 3.1 that θ is a branching point of themechanism iff A ( θ ) + C ( θ ) = 0 . This means that2 gb − ab ch ( θ ) + h − g − b − a + 2 ag ch ( θ ) = 0 , that is ch ( θ ) = a − h a ( g − b ) + g − b a . Corollary 3.3
Suppose that g = b and h (cid:54) = a . Then there is no branchingpoints. roof. If g = b and h (cid:54) = a , then A ( θ ) + C ( θ ) = h − a (cid:54) = 0 . for each θ . Corollary 3.4
Suppose that g = b and h = a . Then all points are branchingpoints. Proof.
This is clear, since then A ( θ ) + C ( θ ) = 0for each θ . Remark.
The deeper investigation of branching points will be one of thetopics of our next article.
Let φ denote the angle between the vectors −→ AO and −→ AB (see Figure 3). • •• • O CA B (cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23) (cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39) (cid:82) (cid:82) φ ga h b
Figure 3: The coupler angle12 heorem 3.5 If φ denotes the coupler angle, then φ = artanh (cid:18) b sh ( ψ ) − a sh ( θ ) g + b ch ( ψ ) − a ch ( θ ) (cid:19) − θ + π Proof.
Since φ was the coupler angle, hence θ + φ − π is the angle to AB fromthe x -axis of F . Consequently we can write the coordinates of B in terms of φ as B = ( a ch ( θ ) + h ch ( θ + φ − π ) , a sh ( θ ) + h sh ( θ + φ − π )) . (9)If we equate the two forms (5) and (9) for B , we obtain the following loopequations of the four-bar linkage: a ch ( θ ) + h ch ( θ + φ − π ) = g + b ch ( ψ ) a sh ( θ ) + h sh ( θ + φ − π ) = b sh ( ψ )Therefore, for a given value of the driving crank θ , ch ( θ + φ − π ) andsh ( θ + φ − π ) are given bych ( θ + φ − π ) = g + b ch ( ψ ) − a ch ( θ ) h and sh ( θ + φ − π ) = b sh ( ψ ) − a sh ( θ ) h . Hence th ( θ + φ − π ) = b sh ( ψ ) − a sh ( θ ) g + b ch ( ψ ) − a ch ( θ )and Theorem 3.5 follows. Remark.
Notice that we obtain a unique value for φ associated with eachof the two solutions for the output angle ψ .13 •• • O CA B (cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23) (cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39) (cid:21) (cid:21) ζga h b
Figure 4: The transmission angle
Let ζ denote the angle between the coupler and the driven crank at B , the transmission angle of the linkage (see Figure 4). Theorem 3.6 If ζ is the transmission angle of the linkage, then ζ = arch (cid:18) − g − a + h + b + 2 ag ch ( θ )2 bh (cid:19) . Proof.
Let d := (cid:107) −→ AC (cid:107) .To determine ζ in terms of θ , equate the Minkowskian cosine laws (The-orem 2.4) for the diagonal AC for the triangles (cid:52) COA and (cid:52)
ABC . Since ζ is the interior angle at B , we have d = g + a − ag ch ( θ ) = h + b − bh ch ( ζ ) . The result follows
In this subsection, we study the motion of the coupler by analyzing the curvetraced by a point on the coupler link. We get the parametrized equation ofthis curve, the coupler curve from the kinematics equations of the drivingRR chain (see Figure 5). 14 •• •
O CA BX (cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23)(cid:23) (cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:39)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7) (cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43) θ λ µr sga h b
Figure 5: The coupler curve
Theorem 3.7
The curve traced by any point of the coupler link of a Minkowskianplanar four-bar linkage is algebraic, of sixth degree.
Proof.
Let x = ( x, y ) T be the coordinates of a coupler point in the frame M located at A with its x -axis along AB .Let X = ( X, Y ) T be the coordinates of a coupler point in the frame F .We obtain the algebraic equation of the coupler curve by defining thecoordinates of X = ( X, Y ) T from two points of view. Let the coupler triangle (cid:52) XAB have length r and s given by r = (cid:107) −−→ AX (cid:107) = (cid:112) x − y and s = (cid:107) −−→ BX (cid:107) = (cid:112) ( x − h ) − y . If λ is the angle to AX in F and µ is the angle to BX (see Figure 5), thenwe have by definition that −−→ AX = ( r ch ( λ ) , r sh ( λ ))and −−→ BX = ( s ch ( µ ) , s sh ( µ )) .
15f we substitute into the identities (cid:107) −→ OA (cid:107) = a and (cid:107) −−→ CB (cid:107) = b andrearrange these equations, we obtain( X − s ch ( µ ) − g ) − ( Y − s sh ( µ )) = h (10)and ( X − r ch ( λ )) − ( Y − r sh ( λ )) = a . (11)If we expand equations (10) and (11), we get2 Xs ch ( µ ) − Y s sh ( µ ) − gs ch ( µ ) − X + 2 Xg + Y − g + h − s = 0 (12)and 2 Xr ch ( λ ) − Y r sh ( λ ) − X + Y + a − r = 0 . (13)Let γ := µ − λ and substitute µ = λ + γ into the equation (12).If we rearrange these equations we get A ( X, Y )ch ( λ ) + B ( X, Y )sh ( λ ) = C ( X, Y ) ,A ( X, Y )ch ( λ ) + B ( X, Y )sh ( λ ) = C ( X, Y ) , (14)where A ( X, Y ) := 2 Xs sh ( γ ) − Y s ch ( γ ) − gs sh ( γ ) ,B ( X, Y ) := 2 Xs ch ( γ ) − Y s sh ( γ ) − gs ch ( γ ) ,C ( X, Y ) := X − Xg − Y + g − h + s ,A ( X, Y ) := 2 rX,B ( X, Y ) := − rY, and C ( X, Y ) := X − Y − a + r . We can eliminate λ from the equation system (14) by solving linearly for u = ch ( λ ) and v = sh ( λ ). Then we can impose the condition u − v = 1.This yields to( C ( X, Y ) B ( X, Y ) − C ( X, Y ) B ( X, Y )) − ( A ( X, Y ) C ( X, Y ) − A ( X, Y ) C ( X, Y )) −− ( A ( X, Y ) B ( X, Y ) − A ( X, Y ) B ( X, Y )) = 0 . Notice that A i ( X, Y ) and B i ( X, Y ) are linear in the coordinates X and Y ,and C i ( X, Y ) are quadratic. Therefore the equation defines a curve of degree6. 16 .6 The input and the output Crank Angles and theclassification of the 4R linkage
The following Theorem describes the upper and the lower limiting angles.
Theorem 3.8
The upper and the lower limiting angles θ max and θ min thatdefine the range of movement of the input crank,ch ( θ min ) = a + g − ( b + h ) ag and ch ( θ max ) = a + g − ( b − h ) ag . Proof.
The formula (see equation (3) in Theorem 3.1) that defines theoutput angle ψ for a given input angle θ has a real solution if and only if B ( θ ) + C ( θ ) − A ( θ ) ≥
0. We obtain the maximum and the minimum valuesfor θ if we set this condition to zero, which yields the following quadraticequation in ch ( θ )4 a g ch ( θ ) − ag ( g + a − h − b )ch ( θ ) ++(( g + a ) − ( h + b ) )(( g + a ) − ( h − b ) ) = 0 . The roots of this equation are the given equations for ch( θ min ) and ch( θ max ). Remark.
These equations are the cosine laws for the two ways that thetriangle
AOC (cid:52) can be formed with the coupler AB aligned with the outputcrank CB , see Figure 6, Figure 7 and Figure 8.The hyperbolic cosine function does not distinguish between ± θ , so thereare actually two limits for ch( θ max ) above and below OC .If θ max does not exist, then the crank has no lower limit to its movementand it rotated though θ = 0 to reach negative values below the segment OC .Thus ch( θ max ) < a + g − ( b − h ) ag < . • • O CA (cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121) (cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44) • Bθ max ga h b Figure 6: The angles θ min and θ max are the limits to the range of movementof the input link • •• • O CAB (cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44) θ min ga h b − h Figure 7: The angles θ min and θ max are the limits to the range of movementof the input link 18 • •• O CA B (cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121)(cid:121) (cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44)(cid:44) θ min ga bh − b Figure 8: The angles θ min and θ max are the limits to the range of movementof the input linkWe can simplify this to yield( b − h ) − ( a − g ) > , If we factor the difference to two squares, we obtain( − a + g + b − h )( a − g + b − h ) > , that is T T > , where T := − a + g + b − h and T := a − g + b − h .If ch( θ min ) >
1, then both upper and lower limiting angles exist. Theinput crank rocks in one of three separate ranges: (i) θ max ≤ θ ≤ ∞ , (ii) − θ min ≤ θ ≤ θ min or (iii) −∞ ≤ θ ≤ − θ max .It is easy to see that the ch( θ min ) > g − a − b − h )( g − a + b + h ) > , that is, T T >
0, where T := g − a − b − h and T := g − a + b + h .Finally we get the following result: Theorem 3.9
We can identify three types of movement available to the inputcrank of a 4R linkage:1. A crank : T T ≥ and T T ≤ , in which case no θ min , θ max exists,and the input crank rocks though θ = 0 . . A rocker : T T < and T T ≤ , which means that both upper andlower limiting angles exist, and the crank cannot pass though . Instead, itrocks in one of the two separate ranges: (i) θ max ≤ θ ≤ ∞ or (ii) −∞ ≤ θ ≤− θ max .3. A superrocker : T T < and T T > , which means that both upperand lower limiting angles exist. It rocks in one of the three separate ranges:(i) θ max ≤ theta ≤ ∞ , (ii) − θ min ≤ θ ≤ θ min or (iii) −∞ ≤ θ ≤ − θ max . The range of movement of the output crank can be analyzed in the sameway. The limiting positions occur when the input crank OA and the coupler AB become aligned. Theorem 3.10
The limits ψ max and ψ min are defined by the equationsch ( ψ max ) = ( a + h ) − g − b bg and ch ( ψ min ) = ( a − h ) − g − b bg . Remark.
In this case ψ is the exterior angle and changes the sign of thehyperbolic cosine term in the cosine law formula.We find the condition for no lower limit ψ max isch( ψ max ) = ( a + h ) − g − b bg < . Hence ( a + h ) < ( b + g ) that is ( g + b + a + h )( g + b − h − a ) > . T := g + b − h − a and T := g + b + a + h, we get T T > . But clearly T > ψ min ) > a − h − g − b )( a − h + g + b ) >
0, that is T T <
0, where T := − a + h + g + b and T := a − h + g + b .Finally we get the following result. Theorem 3.11
We can identify three types of movement available to theoutput crank of a 4R Minkowskian linkage:1. A crank : T ≥ and T T ≥ , in which case no ψ min , ψ max exists,and the output crank rocks though ψ = 0 .2. A rocker : T < and T T ≥ , which means that both upper andlower limiting angles exist, and the crank cannot pass though . Instead, itrocks in one of the two separate ranges: (i) ψ max ≤ ψ ≤ ∞ or (ii) −∞ ≤ ψ ≤ − ψ max .3. A superrocker : T < and T T < , which means that both upperand lower limiting angles exist. It rocks in one of the three separate ranges:(i) ψ max ≤ ψ ≤ ∞ , (ii) − ψ min ≤ ψ ≤ ψ min or (iii) −∞ ≤ ψ ≤ − ψ max . Hence we can classify a Minkowskian planar 4R linkage by the movementof input and output cranks. For example, a crank-rocker has a rotatableinput link though O and an output link that rocks.Now we define special subclasses of the Minkowskian planar 4R mecha-nisms.We say that a planar 4R is normal , if all the conditions a + b + h ≤ g , a + b + g ≤ h , a + g + h ≤ b and b + g + h ≤ a are satisfied. We say thata planar 4R is strange if it is not normal. This means that there exists onelink, which is longer than the sum of other links’ lengths.We say that a normal planar 4R is rigid , if one of the following conditionsis satisfied: a + b + h = g , a + b + g = h , a + g + h = b and b + g + h = a .21e say that a nonrigid normal planar 4R is irreducible , if T (cid:54) = 0 and T (cid:54) = 0, i.e., a + h (cid:54) = b + g and a + b (cid:54) = g + h .Finally we say that a nonrigid normal planar 4R, which is not irreducible,is a reducible planar 4R.First we investigate the strange mechanisms. Theorem 3.12
Suppose that g > a + b + h . Then the Minkowskian planar4R has a superrocker–crank type. Proof.
Since T > T < T > T > T > Theorem 3.13
Suppose that a > g + b + h . Then the Minkowskian planar4R has a superrocker–superrocker type. Proof.
Since T < T > T < T < T > Theorem 3.14
Suppose that h > g + b + a . Then the Minkowskian planar4R has a crank–superrocker type. Proof.
Since T < T < T < T > T < Theorem 3.15
Suppose that b > g + h + a . Then the Minkowskian planar4R has a crank–crank type. b > g + h + a h > g + b + a g > a + b + h a > g + b + h Table 1: Basic Planar 4R
Strange
Linkage Types
Proof.
Since T > T > T < T > T > Theorem 3.16
Suppose that g = a + b + h . Then the Minkowskian planar4R has a rocker–crank type. Proof.
Since T > T < T = 0, T > T > Theorem 3.17
Suppose that a = g + b + h . Then the Minkowskian planar4R has a rocker–rocker type. Proof.
Since T < T > T < T = 0 and T > Theorem 3.18
Suppose that h = g + b + a . Then the Minkowskian planar4R has a crank–rocker type. roof. Since T < T < T < T > T = 0. Theorem 3.19
Suppose that b = g + h + a . Then the Minkowskian planar4R has a crank–crank type. Proof.
Since T > T > T < T > T > b = g + h + a h = g + b + a g = a + b + h a = g + b + h Table 2: Basic Planar 4R
Rigid
Linkage TypesNow we consider normal, non–rigid reducible linkages.
Theorem 3.20
Suppose that g + b = a + h . Then the Minkowskian planar4R has a crank–crank type. Proof.
Since T = 0, hence T T = 0. Theorem 3.21
Suppose that a + b = g + h . Then the input crank is crank. Proof.
Since T = 0, hence T T = 0.Finally we investigate non–rigid normal Minkowskian planar 4R mecha-nisms. 24 heorem 3.22 For all non–rigid normal Minkowskian planar 4R we have T < , T > and T > . The link lengths a, b, g and h for a 4R chain define the two parameters T , T . Clearly our classification requires only the signs of these parameters. Weassamble here an array for (sgn T , sgn T ) (see Table 3).Linkage Type T T Normal Non–Rigid
Linkage Types
Remark.
It is easy to see from the reversed triangle inequality (Lemma2.2) that the following criterion is analogous to Grashof’s criterion: l + s ≥ p + q, where s is the length of the shortest link, l is the length of the longest linkand p , q are the lengths of the remaining two links. We animated the movement of the Minkowskian planar 4R in Matlab. Inthe following examples we show some pictures from our animation.1. Suppose that a := 1, b := 1, g := 4 and h := 1. Then g > a + b + h ,hence this is a strange Minkowskian planar 4R. We getch( θ min ) = 1 . , ch( θ max ) = 2 . , ch( ψ min ) = − . , ch( ψ max ) = − . .T = g + b − h − a = 3, T = a − g + b − h = − T = g − a − b − h = 1, T = g − a + b + h = 5 and T = a − h + g + b = 5. Since T > T < > T > T >
0, thus this planar 4R has a superrocker–cranktype.The branching points occour at ch ( θ ) = 1 . a := 1 . b := 0 . g := 0 . h := 0 .
4. Then a = g + b + h , hence this is a rigid Minkowskian planar 4R. We getch( θ min ) = 1 , ch( θ max ) ≈ . , ch( ψ min ) = 1 , ch( ψ max ) = 7 .T = g + b − h − a = − . T = a − g + b − h = 0 . T = g − a − b − h = − . T = g − a + b + h = 0 and T = a − h + g + b = 1 .
6. Since T < T > T < T = 0 and T >
0, thus this planar 4R has a rocker–rocker type.Here g = b , but h (cid:54) = a , hence there are no branching points.We show here two pictures from the animation (Figures 13, 14), thecoupler curve (Figure 15) and the transmission curve (Figure 16). In Figure15 the blue curve shows the trajectory of the end of the input crank, theblack curve shows the trajectory of the middle point of the coupler, whilethe red curve shows the trajectory of the end of the input crank3. Suppose that a := 0 . b := 1, g := 2 and h := 2 .
5. Then g + b = a + h ,hence this is a normal, non–rigid reducible Minkowskian planar 4R. We getch( θ min ) = − , ch( θ max ) = 1 , ch( ψ min ) = − . , ch( ψ max ) = 1 . Hence T = g + b − h − a = 0, T = a − g + b − h = − T = g − a − b − h = − T = g − a + b + h = 5 and T = a − h + g + b = 1. Since T = 0, T < T < T > T >
0, thus this planar 4R has a crank–crank type.26
Figure 9: The Minkowskian planar 4R with a = 1, b = 1, h = 1, g = 4 at θ = − π/ Figure 10: The Minkowskian planar 4R with a = 1, b = 1, h = 1, g = 4 at θ = π/ Figure 11: The coupler curve of the Minkowskian planar 4R with a = 1, b = 1, h = 1, g = 4 29 Figure 12: The transmission function of the Minkowskian planar 4R with a = 1, b = 1, h = 1, g = 4 30 Figure 13: The Minkowskian planar 4R with a = 1 . b = 0 . h = 0 . g = 0 . θ = − π/ Figure 14: The Minkowskian planar 4R with a = 1 . b = 0 . h = 0 . g = 0 . θ = 3 π/ Figure 15: The coupler curve of the Minkowskian planar 4R with a = 1 . b = 0 . h = 0 . g = 0 . Figure 16: The transmission function of the Minkowskian planar 4R with a = 1 . b = 0 . h = 0 . g = 0 . θ ) = −
5, hence there are no branchingpoints..We show here two pictures from the animation (Figures 17, 18), thecoupler curve (Figure 19) and the transmission curve (Figure 20). In Figure19 the blue curve shows the trajectory of the end of the input crank, theblack curve shows the trajectory of the middle point of the coupler, whilethe red curve shows the trajectory of the end of the input crank4. Suppose that a := 0 . b := 1, g := 0 . h := 0 .
5. Then this is anormal, non–rigid irreducible Minkowskian planar 4R. We getch( θ min ) ≈ − . , ch( θ max ) ≈ . , ch( ψ min ) ≈ − . , ch( ψ max ) = − . . Hence T = g + b − h − a = 0 . T = a − g + b − h = 0 . T = g − a − b − h = − . T = g − a + b + h = 1 . T = a − h + g + b = 1 .
8. Since T > T > T < T > T >
0, thus this planar 4R has a crank–crank type.The branching points occour at ch ( θ ) ≈ − . In this article we characterized and classified completely the planar 4R closedchain working on the Minkowskian plane. We derived formulas for the outputcrank angle, the coupler angle and the transmission angle. We found fourbasic types in the classification: the crank-crank, crank–rocker, rocker–crankand rocker–rocker Minkowskian planar 4R mechanisms and we described theMinkowskian Grashof condition.
Acknowledgements.
The first author is grateful to Josef Schicho andMadalina Hodorog for their useful comments.35
Figure 17: The Minkowskian planar 4R with a := 0 . b := 1, g := 2, h := 2 . θ = − π/ Figure 18: The Minkowskian planar 4R with a := 0 . b := 1, g := 2, h := 2 . θ = π/ Figure 19: The coupler curve of the Minkowskian planar 4R with a := 0 . b := 1, g := 2, h := 2 . Figure 20: The transmission function of the Minkowskian planar 4R with a := 0 . b := 1, g := 2, h := 2 . Figure 21: The Minkowskian planar 4R with a := 0 . b := 1, g := 0 . h := 0 . θ = − π/ Figure 22: The Minkowskian planar 4R with a := 0 . b := 1, g := 0 . h := 0 . θ = π/ Figure 23: The coupler curve of the Minkowskian planar 4R with a := 0 . b := 1, g := 0 . h := 0 . Figure 24: The transmission function of the Minkowskian planar 4R with a := 0 . b := 1, g := 0 . h := 0 . eferences [1] G. S. Birman; K. Nomizu, Trigonometry in Lorentzian geometry. Amer. Math. Monthly (1984), no. 9, 543–549.[2] F. Catoni, R. Cannata, V. Catoni, P. Zampetti, Hyperbolic trigonome-try in two-dimensional space-time geometry. Nuovo Cimento Soc. Ital.Fis. B (2003), no. 5, 475–492.[3] F. Catoni, R. Cannata, V. Catoni, P. Zampetti, Two-dimensional hy-percomplex numbers and related trigonometries and geometries Adv.Appl. Clifford Algebr. (2004), no. 1, 47–68.[4] A. A. Ergin, On the 1-parameter Lorentzian motions, Comm. Fac. Sci.Univ. Ankara Ser. A Math. Statist. 40 (1991), 59–66.[5] P. Fjelstad; S. G. Gal, Two-dimensional geometries, topologies,trigonometries and physics generated by complex-type numbers.
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