Statics Basilio Bona DAUIN Politecnico di Torino 2009 Basilio - - PowerPoint PPT Presentation

Statics

Basilio Bona

DAUIN – Politecnico di Torino

2009

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 1 / 16

Statics - An Introduction

Statics studies the equilibrium or equivalence relations between “cartesian” forces and torques, and their counterparts at the robot joints. Cartesian Space The 3×1 force and torque vectors f and N f =   fx fy fz   N =   Nx Ny Nz   acting on the manipulator TCP are collected in a single vector called cartesian generalized force, or briefly cartesian force F F =

• f

N

• Basilio Bona (DAUIN – Politecnico di Torino)

Statics 2009 2 / 16

Statics - An Introduction

Joint Space We call joint generalized forces the vector τ τ =   τ1 . . . τn   whose elements are the active forces or torques applied by motors to the

• joints. With active we denote only those components of each force/torque

along the motion axis; taking into account the DH conventions, these active components are: τi = kT

i−1Ni−1,i

prismatic joint τi = kT

i−1fi−1,i

rotation joint All the other components are acting on the constraints and do no produce work.

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 3 / 16

Virtual Works Principle

Work is a real quantity obtained as the vector product of a force/torque and a linear/angular displacement L = fTx L = NTα Taking into account generalized forces, we obtain the “cartesian” work at the TCP Lp = FTp and at the joints Lj = τTq

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 4 / 16

Virtual Works Principle

The lavoro virtuale is computed considering only the virtual displacements, i.e., those geometrical displacements that are compatible (or admissible) with the constraints acting on the robot structure. The cartesian virtual displacements are called δp, while the joint virtual displacements are called δq Hence, the cartesian virtual work will be δLp = FTδp while the joint virtual work will be δLj = τTδq

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 5 / 16

Virtual Works Principle

The Virtual Works Principle states that in a static equilibrium condition the virtual work of the generalized cartesian forces equals the virtual work

• f the generalized joint forces

δLp = δLj ⇔ τTδq = FTδp Recalling the kinematic relation ˙ p = J ˙ q, from its differential equivalent dp = Jdq it follows ⇒ δp = Jδq So we obtain δLp = δLj ⇔ τTδq = FTJδq Since the last identity must be true for any δq, it must follow that τT = FTJ

• r, transposing the result

τ = JTF

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 6 / 16

Kineto-Static Relation

We have thus obtained a kineto-static relation τ = JTF that relates the cartesian generalized forces at the TCP with the joint generalized forces. To be precise, the relation should have been written as τ = ±JTF where the ± sign indicates that the work can be assumed positive or negative according to the convention chosen to represent the work done ON the manipulator from the external environment, or the work done BY the manipulator on the external environment.

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 7 / 16

Kineto-Static Relation

To synthesize we assume the following conventions τ = JTF is the equivalence relation between τ e F while τ = −JTF is the equilibrium relation between τ e F The first gives the joint generalized forces that are equivalent to the TCP generalized forces. The second gives the value of the same joint generalized forces that are necessary to compensate for the TCP generalized forces.

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 8 / 16

Image Space and Null Space of J and JT

The kineto-static relation is particularly important for its implications on velocities and generalized force when the manipulator is in a singularity condition. The kinematic singularity appears when detJ = 0. Geometrically, this indicates a rank decrease of the linear transformation ˙ q → ˙ p, represented by the jacobian matrix (square for simplicity) J. Rank falls when the null space dimension of J increases from 0 to a positive value. We recall that, given a linear transformation represented by a square matrix M, the image or range space R(M) is the space spanned by vectors y, such that y = Mx,x ∈ D(M), where D(M) is the transformation domain. The null space or null-space or kernel N (M) of a linear transformation is the space spanned by vectors x, such that 0 = Mx,x ∈ N (M) ⊆ D(M).

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 9 / 16

Image Space and Null Space of J and JT

Given a transformation Mn×n : D(M) → R(M), the following relation holds n = ρ +ν where ρ = dimR and ν = dimN , if ν = 0 transformation is non singular and M has full rank if 0 < ν ≤ ρ transformation is singular and M decreases its rank many times as ν = 0. Moreover the following relations are always true for linear transformations between finite spaces R(M) = N (MT)⊥ N (M) = R(MT)⊥ R(MT) = N (M)⊥ N (MT) = R(M)⊥ (1)

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 10 / 16

Image Space and Null Space of J and JT

If we apply (1) to J and JT, we can establish what follows When J is singular, there are non zero velocities at the joints ˙ qs ∈ N (J) that produce zero cartesian velocities. In this case there are moments τs ∈ N (J) that belong to the

• rthogonal complement of R(JT) and that cannot be balanced by

any generalized cartesian force F ∈ R(J). When JT ` e is singular, we can apply to the TCP a generalized force Fs ∈ N (JT) that do not requires any joint force to be compensated, i.e., τs = JTFs = 0.

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 11 / 16

Example

Figure: A planar 2R manipulator.

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 12 / 16

Example

The kinematic equations of the 2R manipulator are: x = ℓ1c1 +ℓ2c12 y = ℓ1s1 +ℓ2s12 θz = q1 +q2    ⇒ p3 =   ℓ1c1 +ℓ2c12 ℓ1s1 +ℓ2s12 q1 +q2   where p3 includes the only 3 dof of the plane. Jacobian matrix is J =    −ℓ1s1 −ℓ2s12 −ℓ2s12 ℓ1c1 +ℓ2c12 ℓ2c12 ··· ··· 1 1    =

• JL

JA

• Basilio Bona (DAUIN – Politecnico di Torino)

Statics 2009 13 / 16

Example

Looking only at the linear part JL, i.e., considering only the linear motion kinematics and assuming for simplicity that ℓ1 = ℓ2 = 1m, we have JL =

• −s1 −s12

−s12 c1 +c12 c12

• Computing the determinant we obtain

det(JL) = s2 The singularity appears when sin(q2) = 0, that is when q2 = kπ,k = 0,±1,±2,... In singularity conditions we have that

• ˙

x ˙ y

• =
• −s1

c1

• (2˙

q1 + ˙ q2)

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 14 / 16

Example

In singularity, the Jacobian reduces to Js =

• −2s1

−s1 2c1 c1

• JT

s =

• −2s1

2c1 −s1 c1

• If we want to study the null-space or the image space, it is easier to

assume q1 = 0; in this way we can easily compute the following sub-spaces R(J) =

• x = λ
• 1
• ;λ ∈ R
• N (J) =
• x = λ
• 1

−2

• ;λ ∈ R
• R(JT) =
• x = λ
• 2

1

• ;λ ∈ R
• N (JT) =
• x = λ
• 1
• ;λ ∈ R
• Basilio Bona (DAUIN – Politecnico di Torino)

Statics 2009 15 / 16

Example

Figure: Singularity spaces of the Jacobian matrix.

For example, the joint torque τT = k 1 −2 cannot be balanced by any generalized force at the TCP, and the generalized force FT = k 1 does not need any joint bal- ancing torque.

Figure: Which force can balance the two τi at the same time?

Basilio Bona (DAUIN – Politecnico di Torino) Statics 2009 16 / 16