1. ECE 383 / ME 442 Fall 2015 Kris Hauser
Dynamics
ECE 383 / ME 442 Fall 2015
Kris Hauser

2. Issue
Kinematic planning works fine for industrial manipulators: most come with a “go-to” command, and track paths well
But:
How did an engineer make the “go-to” command work in the first place?
They could move faster if timing were considered
What about cars, planes, drones, mobile robots, and humanoid robots?
We need to begin thinking about velocities, forces, inertias, and transmissions… dynamics!

3. Agenda Rigid body dynamics From dynamic equations to simulations
Reference: Murray, Li, and Sastry. MIRM Ch

4. Dynamics
Definition: A description of how a system moves over time in reaction to forces and torques
Distinguished from kinematics, which purely describes states and geometric paths

5. Rigid Body Dynamics
The following can be derived from first principles using Newton’s laws + rigidity assumption
Parameters
CM translation c(t)
CM velocity v(t)
Rotation R(t)
Angular velocity w(t)
Mass m, local inertia tensor HL

6. Rigid body ordinary differential equations
We will express forces and torques in terms of terms of H (a function of R), 𝜔, 𝑣 and 𝜔
𝑓 = 𝑚 𝑣
𝜏 = [𝜔] 𝐻 𝜔 + 𝐻 𝜔
Rearrange…
𝑣 =𝑓/𝑚
𝜔 = 𝐻 −1 (𝜏− 𝜔 𝐻 𝜔 )
So knowing f(t) and τ(t), we can derive c(t), v(t), R(t), and w(t) by solving an ordinary differential equation (ODE)

7. Preliminaries: Newtonian mechanics of a 1D particle
Kinetic energy, ½ mv2
Momentum: derivative of KE w.r.t. velocity, mv
Force: affects the time derivative of momentum, f=ma

8. Newtonian mechanics of a rigid body
Assume a body 𝐵∈ 𝑅 3 has density 𝜌(𝑥)
Velocity 𝑣∈ 𝑅 3 , angular velocity 𝜔∈ 𝑅 3 about some origin point 𝑜∈ 𝑅 3

9. Newtonian mechanics of a rigid body
Assume a body 𝐵∈ 𝑅 3 has density 𝜌(𝑥)
Velocity 𝑣∈ 𝑅 3 , angular velocity 𝜔∈ 𝑅 3 about some origin point 𝑜∈ 𝑅 3
The velocity at a point 𝑥 is 𝑣+𝜔× 𝑥−𝑜

10. Newtonian mechanics of a rigid body
Assume a body 𝐵∈ 𝑅 3 has density 𝜌(𝑥)
Velocity 𝑣∈ 𝑅 3 , angular velocity 𝜔∈ 𝑅 3 about some origin point 𝑜∈ 𝑅 3
The velocity at a point 𝑥 is 𝑣+𝜔× 𝑥−𝑜
The overall KE is 𝐵 𝑣+𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥

11. Newtonian mechanics of a rigid body
KE = 1 2 𝐵 𝑣+𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥 = 1 2 𝐵 𝑣 𝑇 𝑣+2 𝑣 𝑇 𝜔× 𝑥−𝑜 + 𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥

12. Newtonian mechanics of a rigid body
KE = 1 2 𝐵 𝑣+𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥 = 1 2 𝐵 𝑣 𝑇 𝑣+2 𝑣 𝑇 𝜔× 𝑥−𝑜 + 𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥
First term gives 𝑣 𝑇 𝑣 𝐵 𝜌 𝑥 𝑑𝑥 = 1 2 𝑚 𝑣 2
𝑚= 𝐵 𝜌 𝑥 𝑑𝑥 is the mass

13. Newtonian mechanics of a rigid body
KE = 1 2 𝐵 𝑣+𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥 = 1 2 𝐵 𝑣 𝑇 𝑣+2 𝑣 𝑇 𝜔× 𝑥−𝑜 + 𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥
First term gives 1 2 𝑚 𝑣 2
Second term gives 𝑣 𝑇 𝜔× 𝐵 𝑥𝜌 𝑥 𝑑𝑥 −𝑜𝑚
𝐵 𝑥𝜌 𝑥 𝑑𝑥 is m * the center of mass (COM) c
𝑣 𝑇 𝜔× 𝑐−𝑜

14. Newtonian mechanics of a rigid body
KE = 1 2 𝐵 𝑣+𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥 = 1 2 𝐵 𝑣 𝑇 𝑣+2 𝑣 𝑇 𝜔× 𝑥−𝑜 + 𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥
First term gives 1 2 𝑚 𝑣 2
Second term gives 𝑣 𝑇 𝜔× 𝑐−𝑜
Third term gives 𝐵 𝑥−𝑜 𝜔 2 𝜌 𝑥 𝑑𝑥 = 1 2 𝐵 𝜔 𝑇 𝑥−𝑜 𝑇 𝑥−𝑜 𝜔𝜌 𝑥 𝑑𝑥

15. Newtonian mechanics of a rigid body
KE = 1 2 𝐵 𝑣+𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥 = 1 2 𝐵 𝑣 𝑇 𝑣+2 𝑣 𝑇 𝜔× 𝑥−𝑜 + 𝜔× 𝑥−𝑜 2 𝜌 𝑥 𝑑𝑥
First term gives 1 2 𝑚 𝑣 2
Second term gives 𝑣 𝑇 𝜔× 𝑐−𝑜
Third term gives 𝐵 𝑥−𝑜 𝜔 2 𝜌 𝑥 𝑑𝑥 = 1 2 𝐵 𝜔 𝑇 𝑥−𝑜 𝑇 𝑥−𝑜 𝜔𝜌 𝑥 𝑑𝑥 = 1 2 𝜔 𝑇 𝐵 𝑥−𝑜 𝑇 𝑥−𝑜 𝜌 𝑥 𝑑𝑥 𝜔= 1 2 𝜔 𝑇 𝐻𝜔
𝐻 is the inertia tensor about 𝑜

16. Invariants First term gives 1 2 𝑚 𝑣 2 Second term gives 𝑣 𝑇 𝜔× 𝑐−𝑜 𝑚
Third term gives 𝜔 𝑇 𝐻𝜔
Usually, we set o=c, so the second term is eliminated
H changes as the coordinate frame (R,t) moves, so instead we store HL in a reference frame and let H = RHLRT

17. Kinetic energy for rigid body
Rigid body with velocity v, angular velocity w
KE = ½ (m vTv + wT H w)
World-space inertia tensor H = R HL RT T w v
H 0
0 m I w v
1/2

18. Kinetic energy derivatives
𝜕𝐾𝐸 𝜕𝑣 = 𝑚 𝑣
Force 𝑓 = 𝑑 𝑑𝑡 ( 𝜕𝐾𝐸 𝜕𝑣 )= 𝑚 𝑣
𝜕𝐾𝐸 𝜕𝜔 =𝐻𝜔
𝑑 𝑑𝑡 H = [w]H + H[w]
Torque t = 𝑑 𝑑𝑡 𝜕𝐾𝐸 𝜕𝜔 = [w] H w + H 𝜔

19. Summary
𝑓 = 𝑚 𝑣
𝜏 = [𝜔] 𝐻 𝜔 + 𝐻 𝜔
Gyroscopic “force”

20. Force off of COM F x

21. Force off of COM F 𝛿𝑥 x
Consider infinitesimal virtual displacement 𝛿𝑥 generated by F.
(we don’t know what this is, exactly)
The virtual work performed by this displacement is FT𝛿𝑥

22. Generalized torque f
Now consider the equivalent force f, torque τ at COM

23. Generalized torque f 𝛿𝑥 𝛿𝑞
Now consider the equivalent force f, torque τ at COM
And an infinitesimal virtual displacement of R.B. coordinates 𝛿𝑞

24. Generalized torque f 𝛿𝑥 𝛿𝑞
Now consider the equivalent force f, torque τ at COM
And an infinitesimal virtual displacement of R.B. coordinates 𝛿𝑞
Virtual work in configuration space is [fT,τT] 𝛿𝑞

25. Principle of virtual work𝛿𝑥 𝛿𝑞
[fT,τT] 𝛿𝑞= FT 𝛿𝑥
Since 𝛿𝑥=𝐽 𝑞 𝛿𝑞 we have [fT,τT] 𝛿𝑞= FT𝐽 𝑞 𝛿𝑞

26. Principle of virtual work𝛿𝑥 𝛿𝑞
[fT,τT] 𝛿𝑞= FT 𝛿𝑥
Since 𝛿𝑥=𝐽 𝑞 𝛿𝑞 we have [fT,τT] 𝛿𝑞= FT𝐽 𝑞 𝛿𝑞
Since this holds no matter what 𝛿𝑞 is, we have [fT,τT] = FTJ(q),
Or JT(q) F = f τ

27. Simulation
To determine the trajectory, will evolve a system of differential equations over time
dx/dt = f(x)
x(0) = x0
With x=(c,v,R,w) the state of the rigid body
Numerical integration, also known as simulation

28. Numerical integration of ODEs
If dx/dt = f(x) and x(0) are known, then given a step size h,
x(kh)  xk = xk-1 + h f(xk-1)
gives an approximate trajectory for k 1
Provided f is smooth
Accuracy depends on h
Known as Euler’s method

29. Rigid body simulation with Euler integration
At each step:
Compute 𝐻= 𝑅 𝑡 𝐻 𝐿 𝑅 𝑡 𝑇
Sum up all forces at center of mass to obtain ft
Sum up all torques about center of mass and add [ 𝜔 𝑡 ] 𝐻 𝜔 𝑡 to obtain tt
Integrate 𝑣 𝑡+1 ← 𝑣 𝑡 + 𝑓 𝑡 /𝑚Δ𝑡
Integrate 𝜔 𝑡+1 ← 𝜔 𝑡 + 𝐻 −1 𝜏 𝑡 Δ𝑡
Integrate 𝑐 𝑡+1 ← 𝑐 𝑡 + 𝑣 𝑡 Δ𝑡
Integrate 𝑅 𝑡+1 ← 𝑅 𝑒𝑚 ( 𝜔 𝑡 Δ𝑡) 𝑅 𝑡
Note: accumulating numerical errors will cause Rt to drift away from being a perfect rotation matrix

30. Integration errors Lower error with smaller step size
Consider system whose limit cycle is a circle
𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 = −𝑦 𝑥
Euler integrator diverges for all step sizes!
Better integration schemes are available
(e.g., Runge-Kutta methods, implicit integration, adaptive step sizes, energy conservation methods, etc)
Beyond the scope of this course

31. Dynamic equations x(t): state trajectory
(a function from a real number indicating time to a state vector which may include both configuration and velocity)
Uncontrolled dynamic equation:
𝑑𝑥 𝑑𝑡 𝑡 =𝑓(𝑥 𝑡 ,𝑡)
An ordinary differential equation (ODE)

32. From second-order ODEs to first-order ODEs
Let 𝑥(𝑡)=(𝑝(𝑡),𝑣(𝑡)) with 𝑣(𝑡)= 𝑑𝑝 𝑡 𝑑𝑡
Then
𝑑𝑥 𝑑𝑡 𝑡 = 𝑑𝑝 𝑑𝑡 (𝑡) 𝑑𝑣 𝑑𝑡 (𝑡)

33. From second-order ODEs to first-order ODEs
Let 𝑥(𝑡)=(𝑝(𝑡),𝑣(𝑡)) with 𝑣(𝑡)= 𝑑𝑝 𝑡 𝑑𝑡
Then
𝑑𝑥 𝑑𝑡 𝑡 = 𝑑𝑝 𝑑𝑡 (𝑡) 𝑑𝑣 𝑑𝑡 (𝑡) = 𝑣 𝑡 𝐺 𝑚 =𝑓 𝑥 𝑡 ,𝑡
Here G is the gravity vector 0 −𝑔
If p is d dimensional, x is 2d-dimensional

34. From time-dependent to time-independent dynamics
If 𝑑𝑥 𝑑𝑡 (𝑡)=𝑓(𝑥 𝑡 ,𝑡)
Let y(𝑡)=(𝑥 𝑡 ,𝑡)
Then
𝑑𝑦 𝑑𝑡 𝑡 = 𝑑𝑥 𝑑𝑡 (𝑡) 𝑑𝑡 𝑑𝑡 (𝑡) = 𝑓 𝑥 𝑡 ,𝑡 1 =𝑔 𝑦 𝑡

35. Dynamic equation as a vector field
Can ask:
From some initial condition, along what trajectory does the state evolve?
Where will states from some set of initial conditions end up?
Point (convergence), a cycle (limit cycle), or infinity (divergence)?

36. Controlling dynamic systems
Uncontrolled dynamics:
From initial conditions that include state x0 and time t0, the system evolves to produce a trajectory x(t).
Controlled dynamics:
From initial conditions x0, time t0, and given controls u(t), the system evolves to produce a trajectory x(t)

37. Canonical description of controlled dynamic systems
𝑥 𝑡 =𝑓 𝑥 𝑡 ,𝑢 𝑡
Example: u(t) controls a thruster for a point mass under gravity

38. Open vs. Closed loop Open loop control: Closed loop control :
The controls u(t) only depend on time, not x(t)
E.g., a planned path, sent to the robot
No ability to correct for unexpected errors
Closed loop control :
The controls u(x(t),t) depend both on time and x(t)
Feedback control
Requires the ability to measure x(t) (to some extent)
In either case we have an ODE, once we have chosen the control function

39. Controlled Dynamics -> 1st order time-independent ODE
Open loop case:
If 𝑑𝑥 𝑑𝑡 (𝑡) = 𝑓(𝑥 𝑡 ,𝑢(𝑡)), then let y(t)=(x(t),t)
𝑑𝑦 𝑑𝑡 𝑡 = 𝑓 𝑥 𝑡 ,𝑢 𝑡 1 =𝑔 𝑦 𝑡

40. Controlled Dynamics -> 1st order time-independent ODE
Open loop case:
If 𝑑𝑥 𝑑𝑡 (𝑡) = 𝑓(𝑥 𝑡 ,𝑢(𝑡)), then let y(t)=(x(t),t)
𝑑𝑦 𝑑𝑡 𝑡 = 𝑓 𝑥 𝑡 ,𝑢 𝑡 1 =𝑔 𝑦 𝑡
Closed loop case:
If 𝑑𝑥 𝑑𝑡 (𝑡) = 𝑓(𝑥 𝑡 ,𝑢(𝑥 𝑡 ,𝑡)), then let y(t)=(x(t),t)
𝑑𝑦 𝑑𝑡 𝑡 = 𝑓 𝑥 𝑡 ,𝑢 𝑥 𝑡 ,𝑡 1 =𝑔 𝑦 𝑡

41. Controlled Dynamics -> 1st order time-independent ODE
Open loop case:
If 𝑑𝑥 𝑑𝑡 (𝑡) = 𝑓(𝑥 𝑡 ,𝑢(𝑡)), then let y(t)=(x(t),t)
𝑑𝑦 𝑑𝑡 𝑡 = 𝑓 𝑥 𝑡 ,𝑢 𝑡 1 =𝑔 𝑦 𝑡
Closed loop case:
If 𝑑𝑥 𝑑𝑡 (𝑡) = 𝑓(𝑥 𝑡 ,𝑢(𝑥 𝑡 ,𝑡)), then let y(t)=(x(t),t)
𝑑𝑦 𝑑𝑡 𝑡 = 𝑓 𝑥 𝑡 ,𝑢 𝑥 𝑡 ,𝑡 1 =𝑔 𝑦 𝑡
How do we choose u? A subject for future classes

42. Next class Dynamics of articulated robots
Principles Appendix J + Ch