A New Approach in Plane Kinematics

1. Introduction and Prerequisites

2. Kinematics of a Single Vector in

3. Planar Rigid Body Kinematics

4. Poles of the Planar Motion

5. Relative Motion

6. Bresse Circles

7. Geometric Kinematics

5. Conclusion

1. Introduction and Prerequisites

[1,3]

[1,3,4,15]

.

[4,5,6,15,17]

gets isomorphic to complex numbers by adding a complex structure. This leads us to sym

[1,8,16]

R2

R2

R2

1.1 Symplectic Geometry in

Euclidean vector space we get the standard scalar product (Euclidean structure

and

complex structure with and

is an orthogonal operator and transforms any vector into a skew-orthogonal one [1,8,16]

complex vector space. As a shortcut we will place a tilde '~' symbol over the skew-orthogonal vector vari

in the following, you can replace it with . Yet ap

skew-scalar prod

(symplectic structure )

area of the parallelogram spanned from vector to vector

or oriented area [1,8].

to and from to is equivalent to the standard scalar product of

according to the last rule, i.e. , we can interprete the standard scalar product as a par

Symplectic geometry is an areal type of geometry. So areas are first class citizens instead of lengths and angles

from Euclidean geometry [8].

(Dusa Mc. Duff)

compatible triple

[16].

R2

g

(a1

a2) (b1

b2)

g(a,b)=ab =a b +

1 1 a b .

2 2

J=(0

1

−1

0)J=

−1J=

T−J J =

2−I

=a

~Ja = .(−a2

a1)

x

~Jx

ω

ω(a,b) = b=a

~a b −

1 2 a b .

2 1

a b

=a+b+a

~; =b

~a

~

~−a;a=a

~0 ; b=a

~−a; =b

~a

~b

~ab

b a

~a b

~

b=b

~a

~=a

~b

~ab

R2

Fig. 1: Characteristics of symplectic vectors.

of vector

, where

with respect to any arbi

by (Fig. 1c).

Definition 1.1:

Given two points and , the vector describes the transition from to . Swapping the indices results in

is not needed. If we want to, we are free to define any point on the plane as the origin

.

. Note, that the irrelevance of points as bound vectors contrasts with the rele

1.2 Rotation in the Plane

is an orthonormal matrix with , which can be decomposed into a sym

in . So rotating a vector

can be formulated in a matrix-free notation

(a1

a2)

{e, }

oe

~oeo

a

{e, }e

~(ea

ae

~)

A B rAB A B

r=

BA −r.

AB

O

r+

OA r+

AB r=

BO 0

O

=

AB r−

OB r=

OA r−

BrA

R(θ)R=

TR−1

R(θ) = =(cos θ

sin θ

−sin θ

cos θ)cos θ⋅I+ sin θ⋅J.

e=

ix cos x+isin xR2

θ

Ra = cos θa+ sin θa

~

Lemma 1.2:

The -order derivative of the rotation matrix with respect to angle is

(1.2)

. Deriving equation (1.1) with respect to (denoted by prime) yields

applies. Via and continuing, we finally arrive at rela

Proposition 1.3:

Rotating two vectors by the same angle has no effect on their (skew-)scalar product, i.e. .

1.3 Kinematics in the Plane

planes

invariants because they do not depend on specific points on that plane.

are its strictly separated values, magnitude and unit vector .

with respect to gives

according to (1.2).

kRθ

R=

(k)=

dθk

dR

k

RJ with J=

k k

⎩

⎨

⎧I

J

−I

−J

fork= 0(mod4)

fork= 1(mod4)

fork= 2(mod4)

fork= 3(mod4)

θ

R a =

′−sin θa+ cos θ=a

~JRa

R=

′RJ =JR R =

′′ R J =

′RJ2

(Ra)(Rb) = ab

R2

rre

r=re=r(cos φ

sin φ)

eφ

e=

′=(−sin φ

cos φ)Je =and e=e

~(k)=

dφk

de

k

J e ,

k

Jk

Theorem 2.1:

The time derivative of a vector in the plane is a linear recurrence relation of its previous parallel and or‐

thogonal th time derivative components with respect to its symplectic basis

(2.2)

The first two time derivatives of equation (2.1) are

we formulate it as a linear combination with respect to the sym

.

and by recursion (2.2) leads to

Fig. 2: Trajectory of a moving point

k+ 1

k{e, }e

~

r=

(k+1) e+

a∥

(k+1)

(−a)a˙∥

(k)φ˙⊥

(k)

a⊥

(k+1)

( + a)a˙⊥

(k)φ˙∥

(k)e

~

=r

˙

=r

¨

e+rr˙φ˙e

~

(−r)e+ (2 + r)r¨φ˙2φ˙r˙φ¨e

~

r(k)

{e, }e

~

r=

(k)ae+

∥

(k)a⊥

(k)e

~

r=

(k+1) e+a˙∥

(k)a+

∥

(k)φ˙e

~−a˙⊥

(k)e

~ae

⊥

(k)φ˙

a∥

(k)a⊥

(k)

a=r;

∥

(0)

a= ;

∥

(1) r˙

a=−r;

∥

(2) r¨φ˙2

a=−3−3r;

∥

(3) r

... φ˙2r˙φ¨φ˙

a=−6−12 −(4 + 3 −)r;

∥

(4) r

.... φ˙2r¨φ¨φ˙r˙φ

...φ˙φ¨2φ˙4

a=−10 −30 −(20 + 15 −5 ) ;

∥

(5) r

..... φ˙2r

... φ¨φ˙r¨φ

...φ˙φ¨2φ˙4r˙

−(5 + 10 −10 )r;φ

....φ˙φ

...φ¨φ¨φ˙3

⋮;

a= 0

⊥

(0)

a=r

⊥

(1) φ˙

a= 2 + r

⊥

(2) φ˙r˙φ¨

a= 3 + 3 + ( −)r

⊥

(3) φ˙r¨φ¨r˙φ

... φ˙3

a= 4 + 6 + 4( −) + ( −6 )r

⊥

(4) φ˙r

... φ¨r¨φ

... φ˙3r˙φ

.... φ¨φ˙2

a= 5 + 10 + 10( −) + (5 −30 )

⊥

(5) φ˙r

.... φ¨r

... φ

... φ˙3r¨φ

.... φ¨φ˙2r˙

+( −10 −15 + )rφ

..... φ

...φ˙2φ¨2φ˙φ˙5

⋮

is described by the vector from fixed point to that point. Its veloc

is always tangential to the trajectory (Fig. 2a). Parametric curve is a sufficiently times continu

[15].

is

Remark:

The curve parameter does not necessarily have to be interpreted as time here, but rather as any strictly

monotonically increasing curve parameter.

arc length as curve parameter, an arc element of infinitesimal length at location

[2]

from start location to by integration

Remark:

Rewriting this as and using (2.4) proves to be a unit vector.

to .

Lemma 2.2:

The orthonormal base is the intrinsic Frenet coordinate system for plane curves. The equations

(2.6)

are the Frenet equations of the curve . Term is called curvature [2].

Ar(t)O

(t)r

˙r(t)

R2

r(t) = , t∈(r(t)

1

r(t)

2)R

t

(t) =r

˙((t)r˙1

(t)r˙2)

t

s ds t

ds =dt =r

˙2dt =+r˙1

2r˙2

2dtr˙

s(t)t0t

s(t) = dt

∫t0

t

r

˙2

r(s)

=r

˙=

dt

dr=

ds

dr

dt

ds r′

dt

ds

r=

′r

˙ds

dt r=

′

r˙

r

˙

r′T

{T,N}

=(T′

N′)κ⋅

JT

(0

−1

1

0) (T

N)

r(s)κ

. Differentiating with respect to curve parameter gives us

must be perpendicular to , thus

and normal unit vector . Further

and normal unit vector build a right-handed system (Fig.1b).

Remark:

The Frenet coordinate system for plane curves conforms to the symplectic orthonormal base .

Theorem 2.3:

The center of curvature of the trajectory at moving point is given by

(2.7)

with known velocity and acceleration of point .

We can interprete velocity and normal acceleration of point as the result of an instantanious rota

with angular velocity according to equations (2.3) and Figure 1c, i.e.

can be synthesized and - after multiplying this by - we get

from (2.8), while being allowed to equate

by equation (2.7).

Remark:

Equation (2.7) is the vectorial equivalence to the well known scalar formula for the radius of curvature

[8,11,15,16]

Corrolary 2.4:

The magnitude of the curvature vector in equation (2.7) corresponds to the reciprocal of in Lemma 2.2

(2.9)

T=

21s

T=

ds

d22T⋅T=

′0 .

T′T

T=

′κ(JT)=κ=T

~κN.

κN

N=

′JT =

′−κT.

T N

{T,N} {r, }

′r

~′

A0A

r=

AA0r

~

˙r

¨

r˙2

r

~

˙

r

˙r

¨A

A

A0φ˙

=r

˙and =φ˙r

~A A

0r

¨n−r.φ˙2A A

0

=r

¨nφ˙r

~

˙r

~

˙

=φ˙.

r˙2

r

~

˙r

¨n

r

˙=r

~

˙r

¨nr

~

˙r

¨

A0

A

ρ=−x˙y¨y˙x¨

( + )x˙2y˙22

3

rAA0κ

r=

AA0κ

1

. We substitute into equation (2.7) according to (2.5) and also use

and taking

Multiplication by yields scalar expression (2.9), which serves as a proof for the term in ex

3.1 Rigid Body Frames

Euclidean frames attached to the bodies. So each body is given a lo

here in

frame of reference (Fig.3a). The relation between absolute and relative position coordi

[3,4,6,12,14,15,17,18]

Fig. 3: Moving body frame.

Remark:

Orthogonal matrix is the Lie group of the planar rotation [4]. Matrix and vectors and are kine‐

matic invariants, as they do not depend on the choice of a body point.

=r

˙r, =

′s˙r

¨r+

′′s˙2r′s¨r=

′′ κr

~′

r=

AA0(κ+r)r

~′s˙r

~′s˙2′s¨

r′2s˙2

r

~′s˙

s¨r=

′21

=

AA0.

κ

r

~′r′κ

{x,y}R2

{X,Y}

X=a+xcos θ−ysin θ

Y=b+xsin θ+ycos θ

=

r

(X

Y)

=

+⋅

o

(a

b)

R(θ)

(cos θ

sin θ

−sin θ

cos θ)

q

(x

y)

+⋅

o

(a

b)

Q

(x

y

−y

x)

e(θ)

(cos θ

sin θ)

RSO2R o e

and

and

in Figure 3c to be extended to a symplectic orthogonal basis

3.2 Kinematic Equations

and on it. Position of point

, as well as the relative position of point with respect to

is obtained by

Theorem 3.1:

The motion of a rigid body in the plane with two points and is a combination of the translational motion

of point A and the superimposed rotational motion of B around A, which itself is a composition of a radial and

a tangential angular component and .

The time derivatives of these last two components are a linear recurrence relation of their previous radial

and a tangential th time derivative components with respect to the symplectic basis .

Derivation of order of the equation of motion of point with given motion of point is

(3.5)

The first three time derivatives of equation (3.4) using are

.

. Deriving that equation we obtain the accelerations of or

R Q

(θ) = cos θI+ sin θJ Q =xI+yJ

r=o+ cos θq+ sin θ=q

~o+xe+y.e

~

rAB {r,

AB r

~AB

A B A

O B A

B

r=

Br+

Ar.

AB

A B

ΩrΩt

k+ 1

k{r, }

AB r

~AB

k+ 1 B A

r=

B

(k+1) r−

A

(k+1) ⋅

Ωr

(k+1)

( + ωΩ)Ω

˙r

(k)t

(k)r+

AB ⋅

Ωt

(k+1)

(−ωΩ)Ω

˙t

(k)

r

(k)r

~AB

ω=φ˙

=r

˙B

=r

¨B

=r

...

B

+ωr

˙Ar

~AB

−ωr+r

¨A2AB ω˙r

~AB

−3ωr+ ( −ω)r

...

Aω˙AB ω¨3r

~AB

k

r=

B

(k)r−

A

(k)Ωr+

r

(k)AB Ωt

(k)r

~AB

Ωr

(k)

rAB

k+ 1

r=

B

(k+1) r−

A

(k+1) r−Ω

˙r

(k)AB ωΩ+

r

(k)r

~AB −Ω

˙t

(k)r

~AB ωΩr

t

(k)AB

Remarks:

1. The higher-order angular acceleration values and are kinematic invariants. They result from

and in equations (2.2) and (2.3), respectively, by omitting all summands containing time

derivations of magnitude .

2. Based on the initial values and and the recursion rule (3.5), the radial angular component

is independent of derivative , just as the tangential angular component is independent of

derivative .

[4]. The statements in Remark 2 were men

[7].

Theorem 4.1:

The general planar motion of a rigid body can be interpreted as an instantaneous pure rotation about a specific

point – the instantaneous center of velocity . The location of that velocity pole observed from body point is

(4.1)

Substituting point in the first equation of (3.6) by the pole

resolves to the velocity pole's location (4.1).

Remark:

The instantaneous center of velocity is also called velocity pole or simply pole . Due to its high importance in

plane kinematics, we will refer to the velocity pole location as vector for the sake of further intensive use.

currently has no velocity, its acceleration is not zero. Pole accel

Ω=−1 ;

r

(0)

Ω= 0 ;

r

(1)

Ω=ω;

r

(2) 2

Ω= 3 ω;

r

(3) ω˙

Ω= 4 ω+ 3 −ω;

r

(4) ω¨ω˙2 4

Ω= 5 ω+ 10 −10 ω;

r

(5) ω

... ω¨ω˙ω˙3

⋮;

Ω= 0

t

(0)

Ω=ω

t

(1)

Ω=

t

(2) ω˙

Ω=−ω

t

(3) ω¨3

Ω=−6ω

t

(4) ω

... ω˙2

Ω=−10 ω−15 ω+ω

t

(5) ω

.... ω¨2ω˙2 5

⋮

Ωr

(k)Ωt

(k)

a∥

(k)a⊥

(k)

r

Ωr

(0) Ωt

(0) Ωr

(k)

ω(k−1) Ωt

(k)

ω(k−2)

P A

r=

AP ω

r

~

˙A

B P

=

˙P+r

˙Aω⋅=r

~AP 0

P

p

p

¨

P

=p

¨−r

¨Aωr+

2AP .ω˙r

~AP

pole displacement velocity [5].

Theorem 4.2:

In a planar motion the pole displacement velocity is directed perpendicular to the pole's acceleration .

(4.3)

We take equation (4.1), apply orthogonal operator and substitute vector

Reusing

as the pole displacement velocity results in

Remark:

The direction of the pole displacement velocity coincides with the direction of the pole path tangent , whereas

the pole acceleration coincides with the direction of the pole path normal .

Theorem 4.3:

In a planar motion there is a specific pole for which the acceleration of order is zero. That acceleration

pole of order observed from point is obtained by

(4.4)

where the translational acceleration of some point as well as the body's radial and tangential angular ac‐

celeration state and of order is known.

We substitute point in equation (3.4) by acceleration pole and demand it to zero.

and the latter by

This resolves to the location of the acceleration pole (of or

) observed from point in equation (4.4).

u

P

u p

¨

u=ω

p

~

¨

J r =

AP p−rA

=r

˙A−ω=r

~AP ω(−r

~A)p

~

=r

¨A−+ω˙r

~AP ω−r

~

˙Aω.p

~

˙=r

~

˙AωrAP

p

˙u

=r

¨A−+ω˙r

~AP ωr−

2AP ω.u

~

=p

¨−ω.u

~

ut

p

¨n

Pkk

k A

r=

APkΩ+Ω

r

(k)2

t

(k)2

Ω ⋅ r+Ω

r

(k)

A

(k)

t

(k)r

~A

(k)

rA

(k)A

Ωr

(k)Ωt

(k)k

B Pk

r=

Pk

(k)r−

A

(k)Ω ⋅

r

(k)r+

APkΩ ⋅

t

(k)=r

~APk0

rA

(k)

Ωr

(k)Ωt

(k)

+

r=Ω ⋅ r− Ω ⋅ ∣ ⋅ Ω

A

(k)

r

(k)APkt

(k)r

~APkr

(k)

=Ω ⋅ +Ω ⋅ r∣ ⋅ Ωr

~A

(k)

r

(k)r

~APkt

(k)APkt

(k)

Ωr+

r

(k)

A

(k)Ω=

t

(k)r

~A

(k)(Ω+

r

(k)2 Ω)r.

t

(k)2 APk

k A

Remark:

The pole equations (4.4) and (4.5) are not kinematic invariants, as they depend on the choice of the body point

. This can be changed by simply replacing with the velocity pole , which we will do below.

Fig. 4: Relative motion of three bodies.

, and move relative to each other. We want to consider frame

5.1 Relative Kinematics

with respect to frame is denoted by (Figure 4a). When changing perspec

.

Theorem 5.1:

In the case of planar relative motion of three frames , , , their angles, angular velocities and angular acceler‐

ations of order obey the relationship [9]

(5.1)

Angle in Figure 2a can be expressed via the sum

If we derive this several times with respect to time, we obtain by substituting

r=

AP1,r=

ω

r

~

˙A

AP2,r=

ω+

4ω˙2

ω+

2r

¨Aω˙r

~

¨A

AP3,

(3 ω) + ( −ω)ω˙2ω¨3 2

3ω+ ( −ω)ω˙r

...

Aω¨3r

~

...

A



r=

AP4, …

(4 ω+ 3 −ω) + ( −6ω)ω¨ω˙2 4 2 ω

... ω˙2 2

(4 ω+ 3 −ω) + ( −6ω)ω¨ω˙2 4 r

....

Aω

... ω˙2r

~

....

A

A A P

i j k i

i j φij

φ=

ji −φij

i j k

m

ω+

ij

(m)ω+

jk

(m)ω=

ki

(m)0

φki φ=

ki φ+

ji φkj

+

ij φ+

jk φ=

ki 0 . ω=φ˙

φ+

ij φ+

jk φ=

ki 0

ω+

ij ω+

jk ω=

ki 0

+ω˙ij +ω˙jk =ω˙ki 0

⋮

.

Remark:

Read the term as "velocity of point fixed to frame with respect to frame ".

of point A is equal to the sum of the velocities

Thus

Remark:

Observe the cyclic permutation of indices in equations (5.1), (5.2) and (5.3).

5.2 Relative Poles

fixed to frame with respect to frame we get to the rela

observed from point

Remark:

Since and we verify from expression (5.4) that

we have three relative poles and .

Theorem 5.2: Aronhold-Kennedy

Any three frames in a planar relative motion have three corresponding relative poles which are collinear.

Collinearity of demands

according to (5.3) we get

m

r

˙Aij A i j

r

˙Aki +r

˙Akj

r

˙Aji

=r

˙Aij −.r

˙Aji

+r

˙Aij +r

˙Ajk =r

˙Aki 0.

A i j

rAP ij A

r=

AP ij ωij

r

~

˙Aij

ω=

ji −ωij =r

˙Aij −r

˙Aji r=

AP ij r.

AP ji

i,j,kr,

AP ji rAP kj rAP ki

r,r,r

AP ji AP ki AP kj

(r−

AP ki r)( −

AP ji r

~APki ) =r

~APkj 0

(−

ωki

r

~

˙Aki )( −

ωji

r

~

˙Aji

ωkj

r

˙Akj ) =

ωki

r

˙Aki 0 .

=r

˙Aki +r

˙Aji r

˙Akj

( +

ωki

r

~

˙Aji −

ωki

r

~

˙Akj )( −

ωji

r

~

˙Aji

ωkj

r

˙Akj −

ωki

r

˙Aji ) =

ωki

r

˙Akj 0 .

−

ω ω

ki kj

r

~

˙Aji r

˙Akj +

ω ω

ji kj

r

~

˙Aji r

˙Akj =

ω ω

ji ki

r

~

˙Aji r

˙Akj 0 .

[9].

into a tan

Fig. 5: Tangential and normal acceleration components of a moving body point (a) . Circle equation (b)

by

are then

applies, and correspondingly for points with

applies.

Theorem 6.1: Bresse Circles of order

The locus of points on a moving plane, that have either no tangential or no normal acceleration of order , is a

circle in each case. The respective zero normal circle and the zero tangential circle are named first and second

Bresse circle of order . They are kinematic invariants and their corresponding diameter vectors start from

the pole and point towards the opposite poles and respectively.

(6.1)

Both diameter vectors are perpendicular to each other. The acceleration of the pole as well as the

body's radial and tangential angular acceleration state and of order are defined in Theorem 3.1.

We query moving points that, observed from pole , satisfy either the condition

Equation (4.1) gives us

zero normal acceleration condition yields

=r

~

˙Aji r

˙Akj ω ω ω

ji ki kj

ω−ω+ω

ji ki kj 0 .

k

r

˙v

r=

T

(k)v,r=

v2

vr(k)

N

(k)

v2

rv

~(k)

v

~

vr =

(k)0

r=v

~(k)0

k

k

k−1

P Nk−1Tk−1

fork>

r=

PNk −1Ωr

(k)

p(k)

r=

PT k−1Ωt

(k)

p

~(k)⎭

⎬

⎫

1

p(k)P

Ωr

(k)Ωt

(k)k

Q P v⋅

Qr=

Q

(k)0

⋅v

~Qr=

Q

(k)0 . v=

Qωr

~PQ

=

Q

(k)p−

(k)Ω ⋅

r

(k)r+

PQ Ω ⋅

t

(k).r

~PQ

zero tangential acceleration condition results in

, where both vectors and

gives the scalar circle equation

the zero normal circle diameter and the zero tangential circle diameter

and from equations (3.7) yields

Fig. 6: First and second order Bresse circles

zero normal acceleration poles and the zero tangential acceleration poles

Remark:

The velocity pole is a point on all Bresse circles, even though it generally has an acceleration unequal to

zero, since it satisfies the conditions (6.2) and (6.3) due to .

r=v

~QQ

(k)(−ωr)(p−

PQ (k)Ω ⋅

r

(k)r+

PQ Ω ⋅

t

(k)) =r

~PQ 0⇒r−

PQ

2r=

Ωr

(k)

p(k)

PQ 0

r=v

~QQ

(k)(ω)(p−r

~PQ (k)Ω ⋅

r

(k)r+

PQ Ω ⋅

t

(k)) =r

~PQ 0⇒r−

PQ

2r=

Ωt

(k)

p

~(k)

PQ 0

a+λ=a

~d a d

d

a

a−

2da = 0

d rPNk−1rPT k−1

Ωn

(k)Ωt

(k)

r=;

PN1ω2

p

¨

r=;

PN23ωω˙

p

...

r=;

PN34ω+ 3 −ωω¨ω˙2 4

p

....

⋮;

r=

PT 1ω˙

p

~

¨

r=

PT 2−ωω¨3

p

~

...

r=

PT 3−6ωω

... ω˙2

p

~

....

⋮

P

p=p

˙(k)p=p

~

˙(k)0 =p

˙0

Remark:

The first Bresse circle of order 1 is called inflection circle and the point on it opposite to the pole is the in‐

flection pole . Diameter vector is named from now on. Having introduced this, pole displacement

velocity (4.3) can be reformulated to

(6.6)

6.1 Higher-Order Acceleration Poles via Bresse Circles

lies at the intersection point of the Bresse circles and . The vector from pole

is orthogonal to the vector from pole to pole , due to right-angled triangle geometry.

Remark:

The acceleration pole must lie at the intersection point of the Bresse circles and since it has nei‐

ther normal nor tangential acceleration of order .

by multiplication with

is this time replaced by pole .

6.2 Ball's Point via Bresse Circles

Definition 6.2:

A point on the moving plane, which has an instantaneous stationary zero curvature of its path, is a point of

undulation and is called Ball's point after Robert Ball [5,6,9,10,11,15].

is collinear with both its acceleration vector

, i.e. . Therefore Ball's point

to

P

WrPN1rP W

u=ω.r

~PW

PkCNk CT k P

PkTk−1Nk−1

PkCNk CT k

k

r=

P Pkλ(−r

~PNk−1) =r

~PTk−1r+

PT k−1μ(r−

PNk−1r)

PT k−1

λ(−r

~PNk−1)r

~PTk−1

λ= .

(−)r

~PNk−1r

~PTk−12

r(−)

PT k−1r

~PNk−1r

~PTk−1

λ=Ω+Ω

r

(k)2

t

(k)2

Ω Ω

r

(k)

t

(k)

A P

r

˙r

¨

r

... =r

~

˙r

¨=r

~

˙r

... 0U

N2N1

by .

7.1 Conjugate Points

body has an instantaneous center of velocity during motion, the trajectory of some point

instant center of curvature

Definition 7.1:

A point on a moving plane and the instant center of curvature of its trajectory with respect to some

other plane are called conjugate points. Together with the pole they lie on a common straight line - the pole

ray [9,10,15].

in equation (4.1) and in equation (7.1) are directed orthogonal to

suffice as a proof of the collinearity of , and

and

and Equating both, removing

or removing via multiplication by instead, leads to [9]

7.2 Euler-Savary Equation

[9,10,11].

r=

P U λ(−r

~PN1) =r

~PN2r+

PN1μ(r−

PN1r)

PN2

r=

P U (−

(r−r)

PN1PN22

rr

~PN1PN 2r

~PN1) .r

~PN2

r

¨a

P A

A0

r=

AA0av

~A A

vA

2

v

~A

A A0

P

rAP rAA0v

A P A A .

0

{A,A}

0{B,B}

0

r=

Pr+

Aλr

A AA0r=

Pr+

Bλr.

B BB0λ

r

~BB0λAr

~AA0

r=

AP rand r=

rr

~AA0BB0

rr

~BB0AB

AA0BP r.

rr

~AA0BB0

rr

~AA0AB

BB0

Theorem 7.2: Euler-Savary

The equation of Euler-Savary characterizes the curvature of the trajectory of a point on a moving plane. Its

vectorial form reads

(7.3)

In equation (7.1) of the instant center of curvature at point we substitute velocity

from equation (4.2), resulting in

from equation (6.5) and short

Remark:

The well known equivalent scalar Euler-Savary equation can be directly derived from equation (7.3) [11]

(7.4)

It is called the second form of Euler-Savary equation by Pennestri [15]. For the meaning of the variables herein

also see [11]. In this scalar form, the sign of distance must be taken into account separately. The vector form

(7.3) already contains the necessary directional information.

7.3 Geometric Equation of the Inflection Pole

and

can be synthesized.

[11].

7.4 Geometric Equation of Ball's Point

[11]. It has been derived via vectorization of Bereis' construction. If we know the inflection pole be

and two pairs of conjugate points and , we get Ball's point location via

and Ball's point lie on a common pole ray [10,11].

A

r=

AA0r

r r −r

P W P A P A

2

rP A

2

P A

Av=

Aωr

~PA

a=

Aa−

Pωr+

2P A ω˙r

~PA

r=

AA0(−ωr)

(−ωr)(a−ωr+ )

P A P 2P A ω˙r

~PA

ωr

2P A

P A

a=

Pωr

2P W

ρ= .

Dsin θ−r

r2

r

{A,A}

0{B,B}

0

r=

AA0r;r=

r r −r

P W P A P A

2

rP A

2

P A B B0r

r r −r

P W PB PB

2

rPB

2

PB

rP W

r=

P W r+ 1 −r+ 1

rr

~PA PB

1(PB

2(r r

BB0PB

rPB

2)r

~PA P A

2(r r

AA0P A

rP A

2)r

~PB)

W

W

P{A,A}

0{B,B}

0

r=

P U rwith r=

rP H

2

r r

P W PH

P H P H .

rr

~AA0BB0

(r)r−(r)rr

~PW P A B B0r

~PW PB AA0

H U

7.5 Polodes

moving polode

fixed polode [5,12,14,15,17,18]

in the moving plane.

, time

, allowing us to focus exclusively on the geometric aspects of the mo

[3,12,14,15].

Theorem 7.3:

The equations of the fixed and the moving polode – the latter already rotated into the fixed plane – are

(7.8)

. Differentiating equation (7.7) with respect to (denoted by prime) yields For the ve

applies. Using the derivative according to (1.2) we obtain .

to that result yields

Theorem 7.4:

In the pole both polodes have collinear tangents and rolling contact without sliding.

Differentiating both equations (7.8) for

to the second term again obtains

of a curve equals the magnitude of its corresponding tan

and

Remark:

The first derivative is the geometric pole displacement velocity corresponding to velocity from equation

(4.3). Its magnitude is the diameter of the inflection circle.

p

q

p=o+Rq .

=θ

˙=

dt

dθ1

θ

p

Rq

=o+o

~′

=o

~′

θp=

′o+

′R q .

′

p=

′0 R =

′RJ RJq =R=q

~−o′

−J Rq =o

~′

P

θ

p=

′o+

′.o

~′′

Rq +

′R q =

′.o

~′′

R=

′RJ R q =

′R=q

~−o′

p=

′

Rq =

′

o+

′o

~′′

o+ = p

′o

~′′ ′

Rq =

′p′

ds =x(θ) + y(θ)

′2′2

p′q′

p′u

rP W

[5].

7.6 Higher-Order Derivatives of the Polodes

Theorem 7.5:

The th derivative of the polodes conforming to higher-order geometric accelerations of the pole are:

(7.10)

: The -th derivative of the fixed polode

is

derivative obtains

th derivation while reusing

[13].

k P

p=

(k)o+

(k)o

~(k+1)

Rq =

(k)a(−J)pwith a =

i=0

∑

k

i,ki(k−i)i,k∈

i!(k−i)!

k!N

kp(k)

Rq =

(1) Qwith Q=

1 1 p(1)

θ

Rq =

(2)

=

=

=

Q−R q

1

′ ′ (1)

Q+ (−J)Rq

1

′(1)

Q+ (−J)Q

1

′1

Q2

p+ (−J)p

(2) (1)

Rq =

(3)

=

Q+ (−J)Q

2

′2

Q3

p+ 2(−J)p+ (−J)p

(3) (2) 2 (1)

k−1

Rq =

(k−1)

=

Q+ (−J)Q

k−2

′k−2

Qk−1

p+a(−J)p+a(−J)p+⋯+a(−J)p+ (−J)p

(k−1) 2,k−1(k−2) 3,k−12 (k−3) k−2,k−1

k−2 (2) k−1 (1)

kQk−1

Rq =

(k)

=

=

Q+ (−J)Q

k−1

′k−1

p+a(−J)p+a(−J)p+⋯+ (−J)p+

(k)2,k−1(k−1) 3,k−12 (k−2) k−1 (2)

(−J)p+a(−J)p+⋯+a(−J)p+ (−J)p

(k−1) 2,k−12 (k−2) k−2,k−1

k−1 (2) k(1)

Qk

p+(1+a)(−J)p+ (a+a)(−J)p+⋯+ (a+ 1)(−J)p+ (−J)p

(k)2,k−1(k−1) 2,k−1 3,k−12 (k−2) k−2,k−1

k−1 (2) k(1)

ai,k

=

i,ka+

i−1,k−1ai,k−1

=

i,ki!(k−i)!

k!

's with the powers of

7.6 Curvature Radii of the Polodes

Theorem 7.7:

The equations of the curvature radius vectors of the fixed and moving polodes in their contact point with re‐

spect to the fixed frame are

(7.12)

The curvature radius of the fixed polode

and geometric acceleration into equation (2.7). Similarly, we obtain the radius vector of curva

in the fixed plane by substituting and instead.

7.7 Bottema Invariants

and align their axes by setting

with the

. The fixed and moving frame are then called canonical systems [3,6,14,15,17,18].

and its components we adopt the convention of notation from [3], i.e.

Remark:

The scalar derivatives are referred to as Bottema invariants after O. Bottema, who introduced

them for values that are independent of specific body points in [3]. We have used the term invariants in a less

strict sense.

[14,15] and outlines a valuable histori

[12], Veldkamp [17,18], Freudenstein and others.