The Planar Euler-Savary Equation in Vectorial Notation

Stefan Gössner

Department of Mechanical Engineering, University of Applied Sciences, Dortmund, Germany.

Keywords: Euler-Savary Equation; pole transfer velocity; inflection Pole; inflection circle; rho-curves; cubic curve of stationary curvature; Ball's point; undulation point; geometric kinematics;


The Euler-Savary Equation is discussed from a vectorial point of view. Kinematic properties of a moving link in the plane are taken to derive the inflection circle first. Then eliminating all kinematic values results in the pure geometric equation of Euler-Savary. Proceeding to the advantageous canonical coordinate system the curves of points of constant curvature, the cubic of stationary curvature and Ball's point location are derived.


In mechanism analysis and design knowledge about direction and curvature of point paths on a moving link of kinematic chains proves itself valuable and was intensively studied in the past [1,2,3,4]. In this article the kinematic properties of points on a moving plane are discussed first. From here pure geometric relations will be derived:

  • Inflection Pole and Circle
  • The Curve of Points of Constant Curvature
  • The Cubic of Stationary Curvature
  • Ball's Point

The famous Euler-Savary equation is a central point in the discussion of these properties. This equation is derived here based on a vectorial notation using an orthogonal operator [5,6]. This leads to a general-purpose result, which is independent of the commonly used canonical coordinate system.


Consider a moving plane with known velocity pole PP, rotating with current angular velocity ω\\omega. For the path of some point AA fixed on that plane, we can identify its instant center of curvature A0A\_0.

These two points are called conjugate points, which are lying together with the velocity pole PP on a common line - the pole ray. The velocity vA\\bold v\_A of point AA is directed normal to the pole ray and we get it by

vA=vP+ωr~PAwithvP=0,{\\bold v}\_A = {\\bold v}\_P + \\omega\\,{\\tilde \\bold r}\_{PA} \\quad with \\quad \\bold v\_P = \\bold0\\,,(1)

as the pole has no velocity by definition. The acceleration aA\\bold a\_A of point AA results from derivation of (1) with respect to time

aA=aP+ω˙r~PAω2rPA,{\\bold a}\_A = {\\bold a}\_P + \\dot\\omega\\,{\\tilde \\bold r}\_{PA} - \\omega^2\\,{\\bold r}\_{PA}\\,,(2)

with pole acceleration aP\\bold a\_P and angular acceleration ω˙\\dot\\omega of the plane. Dizioglu suggests [7] a rewrite of equation (1) as

vA=ωr~PA=ω(r~Ar~P){\\bold v}\_A = \\omega\\,{\\tilde \\bold r}\_{PA} = \\omega\\,({\\tilde \\bold r}\_{A}-{\\tilde \\bold r}\_{P})
Fig.1: Point path kinematics

and deriving it again w.r.t. time yields

aA=ω˙r~PA+ωr~˙Aωr~˙P.{\\bold a}\_A = \\dot\\omega\\,{\\tilde \\bold r}\_{PA} + \\omega\\,{\\dot{\\tilde\\bold r}}\_{A} - \\omega\\,{\\dot{\\tilde\\bold r}}\_{P}\\,.(3)

Herein is r˙A=vA{\\dot\\bold r}\_{A} = {\\bold v}\_A and r˙P{\\dot\\bold r}\_{P} is the pole transfer velocity u\\bold u. Now comparing components of equations (2) and (3) lets us write the pole transfer velocity in terms of the pole acceleration as

u=a~Pω.\\bold u = \\frac{{\\tilde \\bold a}\_P}{\\omega}\\,.(4)

The direction of the pole transfer velocity u\\bold u coincides with the direction of the pole tangent tt, whereas the pole acceleration aP\\bold a\_P coincides with the direction of the pole normal nn.

We can interprete velocity and normal acceleration of point AA additionally to equation (1) as the result of an instantanious rotation about its center of curvature A0A\_0 with angular velocity ωA\\omega\_A, ie.

vA=ωAr~A0AandaAn=ωA2rA0A.{\\bold v}\_A = \\omega\_A\\,{\\tilde \\bold r}\_{A\_0A}\\quad and \\quad {\\bold a}\_{A\_n} = -\\omega\_A^2\\,{\\bold r}\_{A\_0A}\\,.(5)

From these two expressions aAn=ωAv~A{\\bold a}\_{A\_n} = \\omega\_A\\,{\\tilde\\bold v}\_{A} can be synthesized and - after multiplying this by v~A{\\tilde\\bold v}\_{A} - we get to

ωA=aAnv~AvA2.{\\omega}\_A = \\frac{{\\bold a}\_{A\_n}{\\tilde\\bold v}\_{A}}{v\_A^2}\\,.

That term is introduced back into (5), while being allowed to write aAv~A=aAnv~A{\\bold a}\_{A}{\\tilde\\bold v}\_{A} = {\\bold a}\_{A\_n}{\\tilde\\bold v}\_{A} due to the projective character of the dot product. This finally leads us to the location of the center of curvature A0A\_0

rAA0=vA2aAv~Av~A.{\\bold r}\_{AA\_0} = \\frac{v\_A^2}{{\\bold a}\_{A}{\\tilde\\bold v}\_{A}}{\\tilde\\bold v}\_{A}\\,.(6)

Equation (1) in its form rPA=v~Aω{\\bold r}\_{PA} = -\\frac{{\\tilde\\bold v}\_A}{\\omega} and (6) can be used for a proof, that in fact points PP, AA and A0A\_0 are lying on a common line.

Inflection Circle

We now want to have a closer look at points on the moving plane, which are inflection points of their path at current. Those are points at which their curve changes from being concave to convex, or vice versa, so their radius of curvature is instantaniously infinite. Such a point - say EE - does not possess normal acceleration, its acceleration aE\\bold a\_E is rather directed tangential to the curve, as is its velocity vE\\bold v\_E. So the condition of collinearity between those two has to hold

aEv~E=0.{\\bold a}\_{E}\\,{\\tilde\\bold v}\_E = 0\\,.

Reusing equations (1) and (2) gives

(aP+ω˙r~PEω2rPE)(ωrPE)=0({\\bold a}\_P + \\dot\\omega\\,{\\tilde \\bold r}\_{PE} - \\omega^2\\,{\\bold r}\_{PE})\\,(-\\omega\\,{\\bold r}\_{PE}) = 0

and resolving the brackets leads to the quadratics

rPE2aPω2rPE=0.{\\bold r}\_{PE}^2 - \\frac{\\bold a\_P}{\\omega^2}\\bold r\_{PE} = 0\\,.

Completing the square results in

(rPEaP2ω2)2=aP24ω4,({\\bold r}\_{PE} - \\frac{\\bold a\_P}{2\\omega^2})^2 = \\frac{a\_P^2}{4\\omega^4}\\,,

which has the shape (pp0)2=R2(\\bold p - \\bold p\_0)^2 = R^2 of a circle equation in vector notation.

All points on a moving plane, that are inflection points of their path at current, are located on a circle - the inflection circle.

The pole PP is also an element of the inflection circle, as it fulfills the above condition due to vP=0\\bold v\_P = \\bold 0. The point on this circle opposite to the pole is the inflection pole WW (Fig.2). Its location seen from the pole can be extracted as the diameter from the circle equation above as

rPW=aPω2.{\\bold r}\_{PW} = \\frac{\\bold a\_P}{\\omega^2}\\,.(7)

The Euler-Savary Equation

Fig.2: Notations for the Euler-Savary Equation

Substituting rPA=v~Aω{\\bold r}\_{PA} = -\\frac{{\\tilde\\bold v}\_A}{\\omega} from (1) in equation (2) and multiplying that by v~A{\\tilde\\bold v}\_{A} eliminates the term containing the angular acceleration ω˙\\dot\\omega

aAv~A=aPv~AωvA2.{\\bold a}\_A\\,{\\tilde\\bold v}\_{A} = {\\bold a}\_P\\,{\\tilde\\bold v}\_{A} - \\omega\\,v\_A^2\\,.(8)

Multiplication of equation (6) with v~A{\\tilde\\bold v}\_{A} also yields

rAA0v~A=vA4aAv~A,{\\bold r}\_{AA\_0}{\\tilde\\bold v}\_{A} = \\frac{v\_A^4}{{\\bold a}\_{A}{\\tilde\\bold v}\_{A}}\\,,

which can be resolved for aAv~A{\\bold a}\_{A}{\\tilde\\bold v}\_{A} and introduced to (8)

aPv~A=vA4rAA0v~A+ωvA2.{\\bold a}\_P\\,{\\tilde\\bold v}\_{A} = \\frac{v\_A^4}{{\\bold r}\_{AA\_0}{\\tilde\\bold v}\_{A}} + \\omega\\,v\_A^2\\,.

Now reuse of the terms v~A=ωrPA{\\tilde\\bold v}\_A = -\\omega\\,{\\bold r}\_{PA} and aP=ω2rPW\\bold a\_P = \\omega^2 {\\bold r}\_{PW} from equation (7) helps to remove all kinematic values, finally resulting in the vectorial Euler-Savary equation

rPWrPA=rPA2(rPA2rAA0rPA+1).{\\bold r}\_{PW}{\\bold r}\_{PA} = r\_{PA}^2\\left(\\frac{r\_{PA}^2}{{\\bold r}\_{AA\_0}{\\bold r}\_{PA}} + 1\\right)\\,.(9)

The Euler-Savary equation (9) associates conjugate points of a moving plane with the relative location of velocity pole and inflection pole in a pure geometrical form.

The intersection point AWA\_W of the pole ray with the inflection circle (Fig. 2) is found to be

rPAW=rPA(rPA2rAA0rPA+1),{\\bold r}\_{PA\_W} = {\\bold r}\_{PA}\\left(\\frac{r\_{PA}^2}{{\\bold r}\_{AA\_0}{\\bold r}\_{PA}} + 1\\right)\\,,(10)

so Euler-Savary equation can be also written as

rPWrPA=rPArPAW.{\\bold r}\_{PW}{\\bold r}\_{PA} = {\\bold r}\_{PA} {\\bold r}\_{PA\_W}\\,.(11)

If we happen to know two pairs of conjugate points A/A0A / A\_0 and B/B0B / B\_0 on the moving plane, we can determine their intersection points AWA\_W and BWB\_W with the inflection circle.

Fig.3: Determination of the Inflection Pole

According to (10) they are

rPAW=rPA(rPA2rAA0rPA+1)andrPBW=rPB(rPB2rBB0rPB+1){\\bold r}\_{PA\_W} = {\\bold r}\_{PA}\\left(\\frac{r\_{PA}^2}{{\\bold r}\_{AA\_0}{\\bold r}\_{PA}} + 1\\right)\\quad and \\quad{\\bold r}\_{PB\_W} = {\\bold r}\_{PB}\\left(\\frac{r\_{PB}^2}{{\\bold r}\_{BB\_0}{\\bold r}\_{PB}} + 1\\right)

and with these we are able to write down the Euler-Savary equation two times

rPWrPA=rPArPAWandrPWrPB=rPBrPBW.{\\bold r}\_{PW}{\\bold r}\_{PA} = {\\bold r}\_{PA} {\\bold r}\_{PA\_W}\\quad and\\quad {\\bold r}\_{PW}{\\bold r}\_{PB} = {\\bold r}\_{PB} {\\bold r}\_{PB\_W}\\,.

Now we can synthesize a geometric vectorial equation for the location of the inflection pole WW from those two

rPW=(rPBrPBW)r~PA(rPArPAW)r~PBr~PArPB.{\\bold r}\_{PW} = \\frac{({\\bold r}\_{PB}{\\bold r}\_{PB\_W}){\\tilde\\bold r}\_{PA} - ({\\bold r}\_{PA}{\\bold r}\_{PA\_W}){\\tilde\\bold r}\_{PB}}{{\\tilde\\bold r}\_{PA}{\\bold r}\_{PB}}\\,.(12)

With the knowledge of two pairs of conjugate points of a moving plane, the location of the reflection pole WW can be determined by equation (12).

Equation (12) is a vectorial alternative to Bobillier's construction of the inflection pole.

If we know the inflection pole's location, we can also calculate to any point AA on the moving plane its center point A0A\_0 of curvature with the help of Euler-Savary equation (9)

rAA0=rPA2rPWrPArPA2rPA.{\\bold r}\_{AA\_0} = \\frac{r\_{PA}^2}{{\\bold r}\_{PW}{\\bold r}\_{PA} - r\_{PA}^2}{\\bold r}\_{PA}\\,.(13)

Equation (13) is the geometric pendant to kinematic equation (6). When the denominator in (13) becomes zero, the radius of curvature of the path of AA is infinite. So the expression

rPWrPArPA2=0 {\\bold r}\_{PW}{\\bold r}\_{PA}- r\_{PA}^2 = 0\\,

is a pure geometric, necessary condition for any point AA located on the inflection circle.

The Curvature of Point Paths

In order to investigate the curvature of the path of a point AA of the moving plane in more detail, we align the x-axis of the reference coordinate system with the pole tangent tt and the y-axis with the pole normal nn.

Fig.4: Notations for discussion of the curvature of a point path

Then we call ρ\\rho the curvature radius, rr and ψ\\psi the polar components of the point location AA, r0r\_0 the distance from pole PP to curvature center A0A\_0, eψ\\bold e\_\\psi the unit vector from PP to AA and DD the diameter of the inflection circle (Fig. 4).

In this canonical system Euler-Savary equation (9) reads

Dreyeψ=r2(r2ρr+1).D\\,r\\,{\\bold e}\_y{\\bold e}\_\\psi = r^2(\\frac{r^2}{\\rho\\,r}+1)\\,.

Using eyeψ=sinψ{\\bold e}\_y{\\bold e}\_\\psi = \\sin\\psi we have

Dsinψ=r(r+ρ)ρ.D\\sin\\psi = \\frac{r(r+\\rho)}{\\rho}\\,.(14)

Introducing r0=r+ρr\_0 = r + \\rho leads to

Dsinψ=rr0r0r,D\\sin\\psi = \\frac{r\\,r\_0}{r\_0 - r}\\,,

which - after inverting - results in the well known scalar Euler-Savary equation

1D=(1r1r0)sinψ.\\frac{1}{D} = (\\frac{1}{r} - \\frac{1}{r\_0})\\sin\\psi\\,.(15)

Herein rr and r0r\_0 have to be interpreted as directed quantities. Equation (15) assumes them to be unidirectional. If they have opposite directions, the minus in the parentheses has to be changed to a plus. Using the vectorial form (9), we have the comfort to work with an arbitrary reference coordinate system and don't need to care about signs. Resolving equation (14) for the curvature radius ρ\\rho, yields

ρ=r2Dsinψr,\\rho = \\frac{r^2}{D\\sin\\psi - r}\\,,(16)

which Freudenstein calls the quadratic form of the Euler-Savary equation [8].

The transformation from an arbitrary user coordinate system to the canonical one aligned with pole tangent and pole normal requires the orientation of rPW=Den\\bold r\_{PW} = D\\,\\bold e\_n\\,. By this convention, the x-axis aligned with the pole tangent is directed opposite to the pole transfer velocity u\\bold u\\,. A vector from pole PP to point AA gets transformed into the canonical system and vice versa by pure rotation. Using the matrix

R(φn)=(e~nen)=(sinφncosφncosφnsinφn)\\bold R(\\varphi\_{n}) = \\begin{pmatrix}-\\tilde\\bold e\_n & \\bold e\_n \\end{pmatrix} = \\begin{pmatrix}\\sin\\varphi\_n & \\cos\\varphi\_n \\\\ -\\cos\\varphi\_n & \\sin\\varphi\_n \\end{pmatrix}

vectors are transformed via

rPA=R(φn)reψandreψ=RT(φn)rPA.\\bold r\_{PA} = \\bold R(\\varphi\_{n})\\,r\\,\\bold e\_\\psi \\quad and \\quad r \\,\\bold e\_\\psi = \\bold R^T(\\varphi\_{n})\\,\\bold r\_{PA} \\,.(17)

The Curve of Points of Constant Curvature

All points of a moving plane, possessing the same radius of curvature ρ\\rho of their paths, are located on a curve called ρ\\rho-curve. We get it by resolving equation (16) for rr

r1,2=ρ2(1±1+4Dρsinψ).r\_{1,2} = \\frac{\\rho}{2}\\left(-1 \\pm \\sqrt{1+\\frac{4D}{\\rho}\\sin\\psi}\\right)\\,.(18)

The curve of corresponding centers of curvature results from substituting r=r0ρr = r\_0 - \\rho in equation (16) and resolving for r0r\_0. Here we yield

r01,2=ρ2(1±1+4Dρsinψ).r\_{0\_{\\,1,2}} = \\frac{\\rho}{2}\\left(1 \\pm \\sqrt{1+\\frac{4D}{\\rho}\\sin\\psi}\\right)\\,.(19)

The ρ\\rho Curves and their ρ0\\rho\_0 pendants of fourbar coupler points with given curvature radii are shown in Fig. 5. Here an abstract length unit ee was used. For comparison, both rocker lengths are 3e3e and 4e4e. We can see clearly, that the corresponding curves (3e3e and 4e4e) are running through the coupler joints and their fixed joints respectively.

Fig.5: ρ- and ρ0-Curves of the Coupler of a Fourbar Mechanism

The plus sign before the square root was used in equations (18) and (19). The alternative curves (negative sign) results in mirrored curves with respect to the pole tangent.

When the given curvature radius goes to infinity, the ρ\\rho-curve approaches the inflection circle.

The Cubic of Stationary Curvature

Some points on the moving plane possess a stationary curvature, i.e. their rate of change of their curvature radius ρ\\rho is zero. In order to get to the cubic of stationary curvature, we need to derive the radius of curvature (16) w.r.t. time and require then

dρdt=0.\\frac{d\\rho}{dt} = 0\\,.

As a result we get the cubic curve of centers of stationary curvature. See [1-4, 7-10] for a more indepth discussion.

sinψcosψr=sinψm+cosψn\\frac{\\sin\\psi\\cos\\psi}{r} = \\frac{\\sin\\psi}{m} + \\frac{\\cos\\psi}{n}(20)

All we need to know about coefficents mm and nn is the fact, that they are constant for each particular position of the plane. So, if we can identify two more points on the curve beneath the pole, that have constant curvature for instance - say AA and BB, we can write down equation (20) two times

sinψAcosψArA=sinψAm+cosψAnandsinψBcosψBrB=sinψBm+cosψBn\\frac{\\sin\\psi\_A\\cos\\psi\_A}{r\_A} = \\frac{\\sin\\psi\_A}{m} + \\frac{\\cos\\psi\_A}{n}\\quad and\\quad \\frac{\\sin\\psi\_B\\cos\\psi\_B}{r\_B} = \\frac{\\sin\\psi\_B}{m} + \\frac{\\cos\\psi\_B}{n}

and then solve for mm and nn, which will result in

m=rArB(sinψAcosψBcosψAsinψB)cosψAcosψB(rBsinψArAsinψB)=L2Mm = \\frac{r\_A r\_B(\\sin\\psi\_A\\cos\\psi\_B - \\cos\\psi\_A\\sin\\psi\_B)}{\\cos\\psi\_A\\cos\\psi\_B(r\_B\\sin\\psi\_A-r\_A\\sin\\psi\_B)} = \\frac{L^2}{M}

n=rArB(sinψAcosψBcosψAsinψB)sinψAsinψB(rAcosψBrBcosψA)=L2Nn = \\frac{r\_A r\_B(\\sin\\psi\_A\\cos\\psi\_B - \\cos\\psi\_A\\sin\\psi\_B)}{\\sin\\psi\_A\\sin\\psi\_B(r\_A\\cos\\psi\_B-r\_B\\cos\\psi\_A)} = \\frac{L^2}{N}

Reintroducing those coefficients into (20), while inverting, yields the cubic equation of all points of the moving plane in polar vector notation

r=L2sinψcosψMsinψ+Ncosψ. r = \\frac{L^2\\sin\\psi\\cos\\psi}{M\\sin\\psi + N\\cos\\psi}\\,.(21)

Fig.6: The Cubic of Stationary Curvature

The plot of that curve in Fig. 6 shows all points of the fourbar coupler with stationary curvature of their paths. The cubic of stationary curvature belongs to a family of curves mathematically termed strophoids. When the denominator in equation (21) approaches zero, i.e. Msinψ+Ncosψ=0M\\sin\\psi + N\\cos\\psi = 0\\,, the radial polar component rr goes to infinity. From that expression we are able to derive the gradient of the asymptote as

tanψ=NM.\\tan\\psi\_\\infty = -\\frac{N}{M}\\,.(22)

With certain poses of the fourbar mechanism one of the constants MM or NN may reach zero value. In such cases the cubic curve has two branches. One of them is a circle and the other is a straight line. The circle obtained for N=0N=0 is then

r=L2Mcosψ.r = \\frac{L^2}{M}\\cos\\psi\\,.

For plotting the curve in practice, one has to calculate the coordinates r=reψ\\bold r = r\\bold\\, e\_\\psi with help of equation (21) for a serie of angles ψ[0...π]\\psi \\in [0 ... \\pi], or better to avoid the discontinuity at the asymptode ψ[ψ+ϵ...ψ+πϵ]\\psi \\in [\\psi\_\\infty+\\epsilon\\quad ... \\quad\\psi\_\\infty+\\pi-\\epsilon] first. Then rotation of the canonical values into the mechanism coordinate system has to be performed by equation (17).

Ball's Point

All points on the cubic possess stationary curvature and all points on the inflection circle are running through an inflection point of their paths, so having instantaneous infinite curvature. Thus the intersection point of the cubic curve with the inflection circle has an important property of stationary infinite curvature of fourth order. That point is called Ball's point and is in fact a point of undulation, where the curvature vanishes, but does not change sign.

The equation of the inflection circle in canonical polar coordinates reads

r=Dsinψ. r = D\\,\\sin\\psi\\,.

Equating it with (21) gives us the angular location of Ball's point UU

tanψu=L2DNDM, \\tan\\psi\_u = \\frac{L^2-DN}{DM}\\,,(23)

which has been marked also in Fig. 6.


A discussion of the path curvature of points on a moving plane using a vectorial approach could be done with comparable low effort. The resulting equations prove valuable for practical engineering applications and for visualization via computer graphics.


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