Dynamics Basilio Bona DAUIN Politecnico di Torino Semester 1, - - PowerPoint PPT Presentation

Dynamics

Basilio Bona

DAUIN – Politecnico di Torino

Semester 1, 2015-16

• B. Bona (DAUIN)

Dynamics Semester 1, 2015-16 1 / 18

Dynamics

Dynamics studies the relations between the 3D space generalized forces acting on a body in dynamical equilibrium and its motion (acceleration, velocity, position). With generalized forces, one refers to both linear forces and angular torques. The generalized forces are of the following type External forces applied by some actuator or other physical cause. Inertial forces. Coriolis, centrifugal and gravitational forces. Elastic forces. Friction and other dissipative forces. Forces generated by electromagnetic interaction.

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Dynamics Semester 1, 2015-16 2 / 18

Moment of a Force

Given a point mass located in a point P represented by r and a force f applied to this point, the moment of the force with respect to the point O is given by r ×f = r ×mdv dt = r × dp dt Applying the derivative rules we obtain r ×f = r× dp dt = d dt (r ×p)− dr dt ×p = d dt (r ×p)−v×p = ˙ h where the last term is always zero, since p = mv and v are always collinear. The vector h = r ×p is called angular momentum and we know that is related by the angular velocity ω by h = Γω.

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Dynamics Semester 1, 2015-16 3 / 18

r ×f is often called torque produced by the force with respect to O, and we can write τ = d dt h = ˙ h Note the analogy f = d dt p = ˙ p = m˙ v τ = d dt h = ˙ h = Γ ˙ ω + ˙ Γω The generalized momentum H is the vector that includes both momenta, while the generalized force F include both forces and torques. Therefore we can write F =

• f

τ

• = d

dt

• p

h

• =

˙ p ˙ h

• = ˙

H

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Dynamics Semester 1, 2015-16 4 / 18

Conclusions

Generalized momentum H(t) H(t) =

• p(t)

h(t)

• =
• Mv(t)

linear momentum Γ(t)ω(t) angular momentum

• Generalized force F(t)

F(t) =

• f(t)

τ(t)

• = ˙

H(t) =

• M˙

v(t) Γ(t) ˙ ω(t)+ ˙ Γ(t)ω(t)

• Recall that

˙ Γω = ω ×[Γω] = S(ω) Γω where Γ is the inertia matrix with respect to the body center-of-mass.

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Dynamics Semester 1, 2015-16 5 / 18

Newton-Euler approach

A rigid body system, composed by n rigid bodies is in dynamic equilibrium when The sum of all forces, including inertial forces, is zero. The sum of all angular torques (with respect to the body center of mass), including inertial torques, is zero. The first condition allows to write the so-called Newton equations, while the second condition allows to write the so-called Euler equations. Both equations are second order differential vector equations, i.e., they are equivalent to three scalar equations, one for each vector component. In a n body system, the number of vector equations is 2n and the number

• f scalar equations is 6n.
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Dynamics Semester 1, 2015-16 6 / 18

Newton equations

Consider the following generic rigid body bi, identified with the index i.

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Newton equations

If: fi−1,i resultant of the forces applied from body i −1 to body i fi+1,i resultant of the forces applied from body i +1 to body i gi local gravity field acceleration vector aci total acceleration of the center-of-mass We can write the i-th Newton equation as: fi−1,i +fi+1,i +migi −miaci = 0

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Dynamics Semester 1, 2015-16 8 / 18

Euler equations

If: Ni−1,i resultant of the torques applied from body (i −1) to body i Ni+1,i resultant of the forces applied from body (i +1) to body i Γi inertia matrix of body i with respect to its center-of-mass rci,i−1 position of the (i −1) BRF origin with respect to center-of-mass rci,i position of the (i +1) BRF with respect to center-of-mass BRF Body Reference Frame We can write the i-th Euler equation as Ni−1,i +Ni+1,i +rci,i−1 ×fi−1,i +rci,i ×fi+1,i

• moment of the forces f

−Γi ˙ ωi −ωi ×Γiωi = 0

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Dynamics Semester 1, 2015-16 9 / 18

Newton-Euler equations

The 2n vectorial Newton-Euler (N-E) equations are difficult to solve, at least symbolically, since the internal constraints between the bodies are to be explicitly considered. These constraints are due to the forces f and torques N transmitted by one body to the other. These constraints play no role in determining the dynamical behavior and the motion laws of the multi-body structure, and their determination is not relevant to the description of the body motion. The constraint forces may be relevant for the design of internal structures, etc., since they are related to stresses and strains in the materials. Lagrange equations are much more immediate and easy to deal with, as they are scalar differential equations, obtained from energy considerations. Nonetheless, from a purely numerical/algorithmic viewpoint N-E equations are easier to solve than Lagrange equations, and several recursive solutions are available in technical literature.

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Dynamics Semester 1, 2015-16 10 / 18

Direct and inverse dynamics

Assuming to have determined a set of generalized coordinates q, the direct dynamics problem consists in the computation of the joint accelerations ¨ q(t) (and hence of the joint velocities ˙ q(t) and positions q(t) by integration) from the knowledge of the command generalized forces Fc(t), and the external generalized forces Fe(t); the initial conditions ˙ q(0) and q(0) at time t = 0. The direct dynamics is at the base of any dynamical simulation algorithm. The inverse dynamics problem consists in the computation of the generalized forces Fc(t) that must be applied by the actuators in order to

• btain the desired dynamic behavior, given the motion variables q(t), ˙

q(t) and ¨ q(t). The solution of this inverse problem is useful for the trajectory planning problems and for the control algorithms implementation.

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Dynamics Semester 1, 2015-16 11 / 18

Example: robotics

Let start from the dynamic model equation that one obtains from a generic robotic structure

∑

j

Hij(q)¨ qj +∑

j ∑ k

hijk(q) ˙ qj ˙ qk +gi(q) = τi The various terms (Hij, hijk and gi) depend on the time configuration q(t) as indicated. To solve them, a large number of computations is necessary; it has been demonstrated that the product number is proportional to n4, i.e., its complexity is O(n4).

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Dynamics Semester 1, 2015-16 12 / 18

Recursive algorithm

To overcome this problem, efficient recursive algorithms have been introduced; they consist of two steps: Forward recursion: we use the kinematic functions, that, given a generic qi allow to compute ˙ qi and ¨ qi

1

given q1, ˙ q1 and ¨ q1, one computes the linear and angular velocities and the acceleration vc1, ω1, ac1, ˙ ω1 of the center-of-mass c1

2

using these values and given q2, ˙ q2 and ¨ q2, one computes the linear and angular velocities and the acceleration vc2, ω2, ac2, ˙ ω2 of the center-of-mass c2;

3

the procedure is repeated until the last body quantities are computed.

Backward recursion: knowing the gravitational forces and the externally applied forces, the command forces are computed backward from the last body link to the first.

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Dynamics Semester 1, 2015-16 13 / 18

Notation - 1

These notations are valid for an open-chain robotic structure. ωℓ

k, αℓ k, aℓ k are respectively the angular velocity, the angular

acceleration and the linear acceleration of the joint k, expressed in Rℓ attached to link ℓ; Rk

k−1 is the rotation matrix from Rk to Rk−1; i.e., xk = Rk k−1xk−1

dℓ

k−1,k is the translation of the origin of Rk−1 from the origin of Rk,

represented in Rℓ; dℓ

k−1,kc is the translation of the origin of Rk−1 to the center-of-mass

• f link k, represented in Rℓ;

fℓ

k−1,k is the constraint force transmitted from link k −1 to link k,

represented in Rℓ; τℓ

k−1,k is the constraint torque transmitted from link k −1 to link k,

represented in Rℓ;

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Notation -2

Fℓ

k,c is the resulting force applied on link k, having its action line

across the center-of-mass of the link k, represented in Rℓ; Tℓ

kc is the resulting torque applied on link k, equal to the moment

with respect to the center-of-mass of the link k, represented in Rℓ; Γℓ

k/c is the inertia matrix of link k with respect to the center-of-mass

• f the link k, represented in Rℓ;

mk is the mass of link k; k = 1T; recall that dk

k,kc = (dk k−1,kc −dk k−1,k) and dk k,kc, dk k−1,kc , dk k−1,k are

constant in Rk,

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Dynamics Semester 1, 2015-16 15 / 18

Recursive algorithm - 1

Recursive Algorithm for a revolute joint manipulator: Initialization ω0

0 = 0,

α0

0 = 0,

a0

0 = −g0

where g0 is the gravitational acceleration vector. Forward recursion (k = 1,...,n). ωk

k = Rk k−1(ωk−1 k−1 + ˙

θkk) (1) αk

k = Rk k−1(αk−1 k−1 +S(ωk−1 k−1) ˙

θkk+ ¨ θkk) (2) ak

k = Rk k−1ak−1 k−1 +S(αk k)dk k−1,k +S(ωk k)S(ωk k)dk k−1,k

(3) Center-of-mass motion computation (k = 1,...,n) Fk

kc = mk

• ak

k +S(αk k)dk k,kc +S(ωk k)S(ωk k)dk k,kc

• (4)

Tk

kc = Γk k/cαk k +S(ωk k)Γk k/cωk k

(5)

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Dynamics Semester 1, 2015-16 16 / 18

Recursive algorithm - 2

Backward recursion (k = n −1,...,0) fk

k−1,k = Rk+1 k

fk+1

k,k+1 +Fk kc

(6) τk

k−1,k = Rk+1 k

τk+1

k,k+1 +S(dk k−1,k)Rk+1 k

fk+1

k,k+1

(7) +S(dk

k−1,kc )Fk kc +Tk kc

(8) Motion Torques τk = kTRk−1

k

τk

k−1,k

This algorithm requires 117n −24 products and 103n −21 sums, that for a manipulator with n = 6, results in 678 products and 597 sums, a much smaller number compared with the Lagrangian approach.

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Dynamics Semester 1, 2015-16 17 / 18

Conclusions

Dynamics equations are essential for modelling and control purposes Modelling is easier to understand adopting the Lagrange energy function Computer program are more efficient if they implement recursive Newton-Euler approach Nonlinear state equations have this form H(q)¨ q(t)+C(q, ˙ q)˙ q(t)+B˙ q+g(q) = τ

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Dynamics Semester 1, 2015-16 18 / 18