# Introduction

Several pieces of the puzzle have come together lately to really demonstrate the power of the scientific python software packages to handle complex dynamic and controls problems (i.e. IPython notebooks, matplotlib animations, python-control, and our software packages: sympy.physics.mechanics and PyDy).

This blog post by Wolfram demonstrates Mathematica's ability to symbolically derive the equations of motion for the n-link pendulum and stabilize it with an LQR controller. This blog post inspired us to replicate the example with all free and open source software.

In this example problem, we derive the equations of motion of an n-link pendulum on a laterally sliding cart and then develop a controller to stabilize it. Balancing a single inverted pendulum is a classic problem that is often a student's first experience with non-linear dynamics and control. The problem here is extended to a general n-link pendulum in which the equations of motion quickly get messy with greater than 2 links.

The diagram below shows the general description of the problem.



In [1]:

from IPython.display import SVG
SVG(filename='n-pendulum-with-cart.svg')




Out[1]:

image/svg+xml



# Setup

This example depends on the following software:

• IPython
• NumPy
• SciPy
• SymPy >= 0.7.6
• matplotlib
• python-control
• avconv

The easiest way to install the Python packages it is to use conda:

$conda install ipython-notebook, numpy, scipy, sympy, matplotlib$ conda install -c https://conda.binstar.org/cwrowley control

avconv should be installed as per its recommended procedure for your operating system.

# Equations of Motion

We'll start by generating the equations of motion for the system with SymPy mechanics. The functionality that mechanics provides is much more in depth than Mathematica's functionality. In the Mathematica example, Lagrangian mechanics were implemented manually with Mathematica's symbolic functionality. mechanics provides an assortment of functions and classes to derive the equations of motion for arbitrarily complex (i.e. configuration constraints, nonholonomic motion constraints, etc) multibody systems in a very natural way. First we import the necessary functionality from SymPy.



In [2]:

import sympy as sm
import sympy.physics.mechanics as me



We can enable mathematical rendering of the resulting equations in the notebook with the following command.



In [3]:

me.init_vprinting()



Now specify the number of links, $n$. I'll start with 5 since the Wolfram folks only showed four.



In [4]:

n = 5



mechanics will need the generalized coordinates, generalized speeds, and the input force which are all time dependent variables and the bob masses, link lengths, and acceleration due to gravity which are all constants. Time, $t$, is also made available because we will need to differentiate with respect to time.



In [5]:

q = me.dynamicsymbols('q:{}'.format(n + 1))  # Generalized coordinates
u = me.dynamicsymbols('u:{}'.format(n + 1))  # Generalized speeds
f = me.dynamicsymbols('f')                   # Force applied to the cart

m = sm.symbols('m:{}'.format(n + 1))         # Mass of each bob
l = sm.symbols('l:{}'.format(n))             # Length of each link
g, t = sm.symbols('g t')                     # Gravity and time



Now we can create and inertial reference frame $I$ and define the point, $O$, as the origin.



In [6]:

I = me.ReferenceFrame('I')  # Inertial reference frame
O = me.Point('O')           # Origin point
O.set_vel(I, 0)             # Origin's velocity is zero



Secondly, we define the define the first point of the pendulum as a particle which has mass. This point can only move laterally and represents the motion of the "cart".



In [7]:

P0 = me.Point('P0')                 # Hinge point of top link
P0.set_pos(O, q[0] * I.x)           # Set the position of P0
P0.set_vel(I, u[0] * I.x)           # Set the velocity of P0
Pa0 = me.Particle('Pa0', P0, m[0])  # Define a particle at P0



Now we can define the $n$ reference frames, particles, gravitational forces, and kinematical differential equations for each of the pendulum links. This is easily done with a loop.



In [8]:

frames = [I]                              # List to hold the n + 1 frames
points = [P0]                             # List to hold the n + 1 points
particles = [Pa0]                         # List to hold the n + 1 particles
forces = [(P0, f * I.x - m[0] * g * I.y)] # List to hold the n + 1 applied forces, including the input force, f
kindiffs = [q[0].diff(t) - u[0]]          # List to hold kinematic ODE's

for i in range(n):
Bi = I.orientnew('B' + str(i), 'Axis', [q[i + 1], I.z])   # Create a new frame
Bi.set_ang_vel(I, u[i + 1] * I.z)                         # Set angular velocity
frames.append(Bi)                                         # Add it to the frames list

Pi = points[-1].locatenew('P' + str(i + 1), l[i] * Bi.x)  # Create a new point
Pi.v2pt_theory(points[-1], I, Bi)                         # Set the velocity
points.append(Pi)                                         # Add it to the points list

Pai = me.Particle('Pa' + str(i + 1), Pi, m[i + 1])        # Create a new particle
particles.append(Pai)                                     # Add it to the particles list

forces.append((Pi, -m[i + 1] * g * I.y))                  # Set the force applied at the point

kindiffs.append(q[i + 1].diff(t) - u[i + 1])              # Define the kinematic ODE:  dq_i / dt - u_i = 0



With all of the necessary point velocities and particle masses defined, the KanesMethod class can be used to derive the equations of motion of the system automatically.



In [9]:

kane = me.KanesMethod(I, q_ind=q, u_ind=u, kd_eqs=kindiffs)  # Initialize the object
fr, frstar = kane.kanes_equations(forces, particles)         # Generate EoM's fr + frstar = 0



The equations of motion are quite long as can been seen below. This is the general nature of most non-simple mutlibody problems. That is why a SymPy is so useful; no more mistakes in algebra, differentiation, or copying hand written equations. Note that trigsimp can take quite a while to complete for extremely large expressions. Below we print $\tilde{M}$ and $\tilde{f}$ from $\tilde{M}\dot{u}=\tilde{f}$ to show the size of the expressions.



In [10]:

sm.trigsimp(kane.mass_matrix)




Out[10]:

$$\left[\begin{matrix}m_{0} + m_{1} + m_{2} + m_{3} + m_{4} + m_{5} & - l_{0} \left(m_{1} + m_{2} + m_{3} + m_{4} + m_{5}\right) \operatorname{sin}\left(q_{1}\right) & - l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) \operatorname{sin}\left(q_{2}\right) & - l_{2} \left(m_{3} + m_{4} + m_{5}\right) \operatorname{sin}\left(q_{3}\right) & - l_{3} \left(m_{4} + m_{5}\right) \operatorname{sin}\left(q_{4}\right) & - l_{4} m_{5} \operatorname{sin}\left(q_{5}\right)\\- l_{0} \left(m_{1} + m_{2} + m_{3} + m_{4} + m_{5}\right) \operatorname{sin}\left(q_{1}\right) & l_{0}^{2} m_{1} + l_{0}^{2} m_{2} + l_{0}^{2} m_{3} + l_{0}^{2} m_{4} + l_{0}^{2} m_{5} & l_{0} l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) \operatorname{cos}\left(q_{1} - q_{2}\right) & l_{0} l_{2} \left(m_{3} + m_{4} + m_{5}\right) \operatorname{cos}\left(q_{1} - q_{3}\right) & l_{0} l_{3} \left(m_{4} + m_{5}\right) \operatorname{cos}\left(q_{1} - q_{4}\right) & l_{0} l_{4} m_{5} \operatorname{cos}\left(q_{1} - q_{5}\right)\\- l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) \operatorname{sin}\left(q_{2}\right) & l_{0} l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) \operatorname{cos}\left(q_{1} - q_{2}\right) & l_{1}^{2} m_{2} + l_{1}^{2} m_{3} + l_{1}^{2} m_{4} + l_{1}^{2} m_{5} & l_{1} l_{2} \left(m_{3} + m_{4} + m_{5}\right) \operatorname{cos}\left(q_{2} - q_{3}\right) & l_{1} l_{3} \left(m_{4} + m_{5}\right) \operatorname{cos}\left(q_{2} - q_{4}\right) & l_{1} l_{4} m_{5} \operatorname{cos}\left(q_{2} - q_{5}\right)\\- l_{2} \left(m_{3} + m_{4} + m_{5}\right) \operatorname{sin}\left(q_{3}\right) & l_{0} l_{2} \left(m_{3} + m_{4} + m_{5}\right) \operatorname{cos}\left(q_{1} - q_{3}\right) & l_{1} l_{2} \left(m_{3} + m_{4} + m_{5}\right) \operatorname{cos}\left(q_{2} - q_{3}\right) & l_{2}^{2} m_{3} + l_{2}^{2} m_{4} + l_{2}^{2} m_{5} & l_{2} l_{3} \left(m_{4} + m_{5}\right) \operatorname{cos}\left(q_{3} - q_{4}\right) & l_{2} l_{4} m_{5} \operatorname{cos}\left(q_{3} - q_{5}\right)\\- l_{3} \left(m_{4} + m_{5}\right) \operatorname{sin}\left(q_{4}\right) & l_{0} l_{3} \left(m_{4} + m_{5}\right) \operatorname{cos}\left(q_{1} - q_{4}\right) & l_{1} l_{3} \left(m_{4} + m_{5}\right) \operatorname{cos}\left(q_{2} - q_{4}\right) & l_{2} l_{3} \left(m_{4} + m_{5}\right) \operatorname{cos}\left(q_{3} - q_{4}\right) & l_{3}^{2} m_{4} + l_{3}^{2} m_{5} & l_{3} l_{4} m_{5} \operatorname{cos}\left(q_{4} - q_{5}\right)\\- l_{4} m_{5} \operatorname{sin}\left(q_{5}\right) & l_{0} l_{4} m_{5} \operatorname{cos}\left(q_{1} - q_{5}\right) & l_{1} l_{4} m_{5} \operatorname{cos}\left(q_{2} - q_{5}\right) & l_{2} l_{4} m_{5} \operatorname{cos}\left(q_{3} - q_{5}\right) & l_{3} l_{4} m_{5} \operatorname{cos}\left(q_{4} - q_{5}\right) & l_{4}^{2} m_{5}\end{matrix}\right]$$



$\tilde{M}$ is a function of the constant parameters and the configuration.



In [11]:

me.find_dynamicsymbols(kane.mass_matrix)




Out[11]:

$$\left\{q_{1}, q_{2}, q_{3}, q_{4}, q_{5}\right\}$$




In [12]:

sm.trigsimp(kane.forcing)




Out[12]:

$$\left[\begin{matrix}l_{0} m_{1} u^{2}_{1} \operatorname{cos}\left(q_{1}\right) + l_{0} m_{2} u^{2}_{1} \operatorname{cos}\left(q_{1}\right) + l_{0} m_{3} u^{2}_{1} \operatorname{cos}\left(q_{1}\right) + l_{0} m_{4} u^{2}_{1} \operatorname{cos}\left(q_{1}\right) + l_{0} m_{5} u^{2}_{1} \operatorname{cos}\left(q_{1}\right) + l_{1} m_{2} u^{2}_{2} \operatorname{cos}\left(q_{2}\right) + l_{1} m_{3} u^{2}_{2} \operatorname{cos}\left(q_{2}\right) + l_{1} m_{4} u^{2}_{2} \operatorname{cos}\left(q_{2}\right) + l_{1} m_{5} u^{2}_{2} \operatorname{cos}\left(q_{2}\right) + l_{2} m_{3} u^{2}_{3} \operatorname{cos}\left(q_{3}\right) + l_{2} m_{4} u^{2}_{3} \operatorname{cos}\left(q_{3}\right) + l_{2} m_{5} u^{2}_{3} \operatorname{cos}\left(q_{3}\right) + l_{3} m_{4} u^{2}_{4} \operatorname{cos}\left(q_{4}\right) + l_{3} m_{5} u^{2}_{4} \operatorname{cos}\left(q_{4}\right) + l_{4} m_{5} u^{2}_{5} \operatorname{cos}\left(q_{5}\right) + f\\- l_{0} \left(g m_{1} \operatorname{cos}\left(q_{1}\right) + g m_{2} \operatorname{cos}\left(q_{1}\right) + g m_{3} \operatorname{cos}\left(q_{1}\right) + g m_{4} \operatorname{cos}\left(q_{1}\right) + g m_{5} \operatorname{cos}\left(q_{1}\right) + l_{1} m_{2} u^{2}_{2} \operatorname{sin}\left(q_{1} - q_{2}\right) + l_{1} m_{3} u^{2}_{2} \operatorname{sin}\left(q_{1} - q_{2}\right) + l_{1} m_{4} u^{2}_{2} \operatorname{sin}\left(q_{1} - q_{2}\right) + l_{1} m_{5} u^{2}_{2} \operatorname{sin}\left(q_{1} - q_{2}\right) + l_{2} m_{3} u^{2}_{3} \operatorname{sin}\left(q_{1} - q_{3}\right) + l_{2} m_{4} u^{2}_{3} \operatorname{sin}\left(q_{1} - q_{3}\right) + l_{2} m_{5} u^{2}_{3} \operatorname{sin}\left(q_{1} - q_{3}\right) + l_{3} m_{4} u^{2}_{4} \operatorname{sin}\left(q_{1} - q_{4}\right) + l_{3} m_{5} u^{2}_{4} \operatorname{sin}\left(q_{1} - q_{4}\right) + l_{4} m_{5} u^{2}_{5} \operatorname{sin}\left(q_{1} - q_{5}\right)\right)\\l_{1} \left(- g m_{2} \operatorname{cos}\left(q_{2}\right) - g m_{3} \operatorname{cos}\left(q_{2}\right) - g m_{4} \operatorname{cos}\left(q_{2}\right) - g m_{5} \operatorname{cos}\left(q_{2}\right) + l_{0} m_{2} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{2}\right) + l_{0} m_{3} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{2}\right) + l_{0} m_{4} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{2}\right) + l_{0} m_{5} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{2}\right) - l_{2} m_{3} u^{2}_{3} \operatorname{sin}\left(q_{2} - q_{3}\right) - l_{2} m_{4} u^{2}_{3} \operatorname{sin}\left(q_{2} - q_{3}\right) - l_{2} m_{5} u^{2}_{3} \operatorname{sin}\left(q_{2} - q_{3}\right) - l_{3} m_{4} u^{2}_{4} \operatorname{sin}\left(q_{2} - q_{4}\right) - l_{3} m_{5} u^{2}_{4} \operatorname{sin}\left(q_{2} - q_{4}\right) - l_{4} m_{5} u^{2}_{5} \operatorname{sin}\left(q_{2} - q_{5}\right)\right)\\l_{2} \left(- g m_{3} \operatorname{cos}\left(q_{3}\right) - g m_{4} \operatorname{cos}\left(q_{3}\right) - g m_{5} \operatorname{cos}\left(q_{3}\right) + l_{0} m_{3} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{3}\right) + l_{0} m_{4} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{3}\right) + l_{0} m_{5} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{3}\right) + l_{1} m_{3} u^{2}_{2} \operatorname{sin}\left(q_{2} - q_{3}\right) + l_{1} m_{4} u^{2}_{2} \operatorname{sin}\left(q_{2} - q_{3}\right) + l_{1} m_{5} u^{2}_{2} \operatorname{sin}\left(q_{2} - q_{3}\right) - l_{3} m_{4} u^{2}_{4} \operatorname{sin}\left(q_{3} - q_{4}\right) - l_{3} m_{5} u^{2}_{4} \operatorname{sin}\left(q_{3} - q_{4}\right) - l_{4} m_{5} u^{2}_{5} \operatorname{sin}\left(q_{3} - q_{5}\right)\right)\\l_{3} \left(- g m_{4} \operatorname{cos}\left(q_{4}\right) - g m_{5} \operatorname{cos}\left(q_{4}\right) + l_{0} m_{4} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{4}\right) + l_{0} m_{5} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{4}\right) + l_{1} m_{4} u^{2}_{2} \operatorname{sin}\left(q_{2} - q_{4}\right) + l_{1} m_{5} u^{2}_{2} \operatorname{sin}\left(q_{2} - q_{4}\right) + l_{2} m_{4} u^{2}_{3} \operatorname{sin}\left(q_{3} - q_{4}\right) + l_{2} m_{5} u^{2}_{3} \operatorname{sin}\left(q_{3} - q_{4}\right) - l_{4} m_{5} u^{2}_{5} \operatorname{sin}\left(q_{4} - q_{5}\right)\right)\\l_{4} m_{5} \left(- g \operatorname{cos}\left(q_{5}\right) + l_{0} u^{2}_{1} \operatorname{sin}\left(q_{1} - q_{5}\right) + l_{1} u^{2}_{2} \operatorname{sin}\left(q_{2} - q_{5}\right) + l_{2} u^{2}_{3} \operatorname{sin}\left(q_{3} - q_{5}\right) + l_{3} u^{2}_{4} \operatorname{sin}\left(q_{4} - q_{5}\right)\right)\end{matrix}\right]$$



$\tilde{f}$ is a function of the constant parameters, configuration, speeds, and the applied force.



In [13]:

me.find_dynamicsymbols(kane.forcing)




Out[13]:

$$\left\{f, q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, u_{1}, u_{2}, u_{3}, u_{4}, u_{5}\right\}$$



# Simulation

Now that the symbolic equations of motion are available we can simulate the pendulum's motion. We will need some more SymPy functionality and several NumPy functions, and most importantly the integration function from SciPy, odeint.



In [14]:

import numpy as np
from numpy.linalg import solve
from scipy.integrate import odeint



First, define some numeric values for all of the constant parameters in the problem.



In [15]:

arm_length = 1. / n                          # The maximum length of the pendulum is 1 meter
bob_mass = 0.01 / n                          # The maximum mass of the bobs is 10 grams
parameters = [g, m[0]]                       # Parameter definitions starting with gravity and the first bob
parameter_vals = [9.81, 0.01 / n]            # Numerical values for the first two
for i in range(n):                           # Then each mass and length
parameters += [l[i], m[i + 1]]
parameter_vals += [arm_length, bob_mass]



Mathematica has a really nice NDSolve function for quickly integrating their symbolic differential equations. We make use of SymPy's lambdify function to do something similar, i.e. to create functions that will evaluate the "full" mass matrix, $M$, and "full" forcing vector, $f$ from $M\dot{x} = f(x, r, t)$ as a NumPy function.



In [16]:

dynamic = q + u    # Make a list of the states
dynamic.append(f)  # Add the input force

M_func = sm.lambdify(dynamic + parameters, kane.mass_matrix_full) # Create a callable function to evaluate the mass matrix
f_func = sm.lambdify(dynamic + parameters, kane.forcing_full)     # Create a callable function to evaluate the forcing vector



To integrate the ODE's we need to define a function that returns the derivatives of the states given the current state and time.



In [17]:

def right_hand_side(x, t, args):
"""Returns the derivatives of the states.

Parameters
----------
x : ndarray, shape(2 * (n + 1))
The current state vector.
t : float
The current time.
args : ndarray
The constants.

Returns
-------
dx : ndarray, shape(2 * (n + 1))
The derivative of the state.

"""
r = 0.0                                  # The input force is always zero
arguments = np.hstack((x, r, args))      # States, input, and parameters
dx = np.array(solve(M_func(*arguments),  # Solving for the derivatives
f_func(*arguments))).T[0]

return dx



Now that we have the right hand side function, the initial conditions are set such that the pendulum is in the vertical equilibrium and a slight initial rate is set for each speed to ensure the pendulum falls. The equations can then be integrated with SciPy's odeint function given a time series.



In [18]:

x0 = np.hstack((0.0,                              # q0
np.pi / 2 * np.ones(len(q) - 1),  # q1...qn+1
1e-3 * np.ones(len(u))))          # u0...un+1




In [19]:

t = np.linspace(0.0, 10.0, num=500)  # Time vector




In [20]:

x = odeint(right_hand_side, x0, t, args=(parameter_vals,))  # Numerical integration



# Plotting

The results of the simulation can be plotted with matplotlib. First, load the plotting functionality.



In [21]:

import matplotlib.pyplot as plt




In [22]:

%matplotlib inline




In [23]:

from IPython.core.pylabtools import figsize
figsize(8.0, 6.0)



The coordinate trajectories are plotted below.



In [24]:

lines = plt.plot(t, x[:, :x.shape[1] / 2])
lab = plt.xlabel('Time [sec]')
leg = plt.legend(dynamic[:x.shape[1] / 2])






And the generalized speed trajectories.



In [25]:

lines = plt.plot(t, x[:, x.shape[1] / 2:])
lab = plt.xlabel('Time [sec]')
leg = plt.legend(dynamic[x.shape[1] / 2:])






# Animation

matplotlib now includes very nice animation functions for animating matplotlib plots. First we import the necessary functions for creating the animation.



In [26]:

from matplotlib import animation
from matplotlib.patches import Rectangle



The following function was modeled from Jake Vanderplas's post on matplotlib animations.



In [27]:

def animate_pendulum(t, states, length, filename=None):
"""Animates the n-pendulum and optionally saves it to file.

Parameters
----------
t : ndarray, shape(m)
Time array.
states: ndarray, shape(m,p)
State time history.
length: float
The length of the pendulum links.
filename: string or None, optional
If true a movie file will be saved of the animation. This may take some time.

Returns
-------
fig : matplotlib.Figure
The figure.
anim : matplotlib.FuncAnimation
The animation.

"""
# the number of pendulum bobs
numpoints = states.shape[1] / 2

# first set up the figure, the axis, and the plot elements we want to animate
fig = plt.figure()

# some dimesions
cart_width = 0.4
cart_height = 0.2

# set the limits based on the motion
xmin = np.around(states[:, 0].min() - cart_width / 2.0, 1)
xmax = np.around(states[:, 0].max() + cart_width / 2.0, 1)

# create the axes
ax = plt.axes(xlim=(xmin, xmax), ylim=(-1.1, 1.1), aspect='equal')

# display the current time
time_text = ax.text(0.04, 0.9, '', transform=ax.transAxes)

# create a rectangular cart
rect = Rectangle([states[0, 0] - cart_width / 2.0, -cart_height / 2],
cart_width, cart_height, fill=True, color='red',
ec='black')

# blank line for the pendulum
line, = ax.plot([], [], lw=2, marker='o', markersize=6)

# initialization function: plot the background of each frame
def init():
time_text.set_text('')
rect.set_xy((0.0, 0.0))
line.set_data([], [])
return time_text, rect, line,

# animation function: update the objects
def animate(i):
time_text.set_text('time = {:2.2f}'.format(t[i]))
rect.set_xy((states[i, 0] - cart_width / 2.0, -cart_height / 2))
x = np.hstack((states[i, 0], np.zeros((numpoints - 1))))
y = np.zeros((numpoints))
for j in np.arange(1, numpoints):
x[j] = x[j - 1] + length * np.cos(states[i, j])
y[j] = y[j - 1] + length * np.sin(states[i, j])
line.set_data(x, y)
return time_text, rect, line,

# call the animator function
anim = animation.FuncAnimation(fig, animate, frames=len(t), init_func=init,
interval=t[-1] / len(t) * 1000, blit=True, repeat=False)

# save the animation if a filename is given
if filename is not None:
anim.save(filename, fps=30, writer="avconv", codec='libx264')



Now we can create the animation of the pendulum. This animation will show the open loop dynamics.



In [28]:

animate_pendulum(t, x, arm_length, filename="open-loop.mp4")







In [29]:

from IPython.display import HTML
html = \
"""
<video width="640" height="480" controls>
<source src="open-loop.mp4" type="video/mp4">
</video>
"""
HTML(html)




Out[29]:



# Controller Design

The n-link pendulum can be balanced such that all of the links are inverted above the cart by applying the correct lateral force to the cart. We can design a full state feedback controller based from a linear model of the pendulum about its upright equilibrium point. We'll start by specifying the equilibrium point and parameters in dictionaries. We make sure to use SymPy types in the equilibrium point to ensure proper cancelations in the linearization.



In [30]:

equilibrium_point = [sm.S(0)] + [sm.pi / 2] * (len(q) - 1) + [sm.S(0)] * len(u)
equilibrium_dict = dict(zip(q + u, equilibrium_point))
equilibrium_dict




Out[30]:

$$\left \{ q_{0} : 0, \quad q_{1} : \frac{\pi}{2}, \quad q_{2} : \frac{\pi}{2}, \quad q_{3} : \frac{\pi}{2}, \quad q_{4} : \frac{\pi}{2}, \quad q_{5} : \frac{\pi}{2}, \quad u_{0} : 0, \quad u_{1} : 0, \quad u_{2} : 0, \quad u_{3} : 0, \quad u_{4} : 0, \quad u_{5} : 0\right \}$$



The KanesMethod class has method that linearizes the forcing vector about generic state and input perturbation vectors. The equilibrium point and numerical constants can then be substituted in to give the linear system in this form: $M\dot{x}=F_Ax+F_Br$. The state and input matrices, $A$ and $B$, can then be computed by left side multiplication by the inverse of the mass matrix: $A=M^{-1}F_A$ and $B=M^{-1}F_B$.



In [31]:

M, F_A, F_B, r = kane.linearize(new_method=True, op_point=equilibrium_dict)




In [32]:

sm.simplify(M)




Out[32]:

$$\left[\begin{array}{cccccccccccc}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & - m_{0} - m_{1} - m_{2} - m_{3} - m_{4} - m_{5} & l_{0} \left(m_{1} + m_{2} + m_{3} + m_{4} + m_{5}\right) & l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) & l_{2} \left(m_{3} + m_{4} + m_{5}\right) & l_{3} \left(m_{4} + m_{5}\right) & l_{4} m_{5}\\0 & 0 & 0 & 0 & 0 & 0 & l_{0} \left(m_{1} + m_{2} + m_{3} + m_{4} + m_{5}\right) & - l_{0}^{2} \left(m_{1} + m_{2} + m_{3} + m_{4} + m_{5}\right) & - l_{0} l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) & - l_{0} l_{2} \left(m_{3} + m_{4} + m_{5}\right) & - l_{0} l_{3} \left(m_{4} + m_{5}\right) & - l_{0} l_{4} m_{5}\\0 & 0 & 0 & 0 & 0 & 0 & l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) & - l_{0} l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) & - l_{1}^{2} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) & - l_{1} l_{2} \left(m_{3} + m_{4} + m_{5}\right) & - l_{1} l_{3} \left(m_{4} + m_{5}\right) & - l_{1} l_{4} m_{5}\\0 & 0 & 0 & 0 & 0 & 0 & l_{2} \left(m_{3} + m_{4} + m_{5}\right) & - l_{0} l_{2} \left(m_{3} + m_{4} + m_{5}\right) & - l_{1} l_{2} \left(m_{3} + m_{4} + m_{5}\right) & - l_{2}^{2} \left(m_{3} + m_{4} + m_{5}\right) & - l_{2} l_{3} \left(m_{4} + m_{5}\right) & - l_{2} l_{4} m_{5}\\0 & 0 & 0 & 0 & 0 & 0 & l_{3} \left(m_{4} + m_{5}\right) & - l_{0} l_{3} \left(m_{4} + m_{5}\right) & - l_{1} l_{3} \left(m_{4} + m_{5}\right) & - l_{2} l_{3} \left(m_{4} + m_{5}\right) & - l_{3}^{2} \left(m_{4} + m_{5}\right) & - l_{3} l_{4} m_{5}\\0 & 0 & 0 & 0 & 0 & 0 & l_{4} m_{5} & - l_{0} l_{4} m_{5} & - l_{1} l_{4} m_{5} & - l_{2} l_{4} m_{5} & - l_{3} l_{4} m_{5} & - l_{4}^{2} m_{5}\end{array}\right]$$




In [33]:

sm.simplify(F_A)




Out[33]:

$$\left[\begin{array}{cccccccccccc}0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & - g l_{0} \left(m_{1} + m_{2} + m_{3} + m_{4} + m_{5}\right) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & - g l_{1} \left(m_{2} + m_{3} + m_{4} + m_{5}\right) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & - g l_{2} \left(m_{3} + m_{4} + m_{5}\right) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & - g l_{3} \left(m_{4} + m_{5}\right) & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & - g l_{4} m_{5} & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]$$




In [34]:

sm.simplify(F_B)




Out[34]:

$$\left[\begin{matrix}0\\0\\0\\0\\0\\0\\-1\\0\\0\\0\\0\\0\end{matrix}\right]$$



Now the numerical $A$ and $B$ matrices can be formed. First substitute numerical parameter values into $M$, $F_A$, and $F_B$.



In [35]:

parameter_dict = dict(zip(parameters, parameter_vals))
parameter_dict




Out[35]:

$$\left \{ g : 9.81, \quad l_{0} : 0.2, \quad l_{1} : 0.2, \quad l_{2} : 0.2, \quad l_{3} : 0.2, \quad l_{4} : 0.2, \quad m_{0} : 0.002, \quad m_{1} : 0.002, \quad m_{2} : 0.002, \quad m_{3} : 0.002, \quad m_{4} : 0.002, \quad m_{5} : 0.002\right \}$$




In [36]:

M_num = sm.matrix2numpy(M.subs(parameter_dict), dtype=float)
F_A_num = sm.matrix2numpy(F_A.subs(parameter_dict), dtype=float)
F_B_num = sm.matrix2numpy(F_B.subs(parameter_dict), dtype=float)




In [37]:

A = np.linalg.solve(M_num, F_A_num)
B = np.linalg.solve(M_num ,F_B_num)




In [38]:

print(A)




[[  0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   1.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00]
[  0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   1.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00]
[  0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
1.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00]
[  0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   1.00000000e+00   0.00000000e+00   0.00000000e+00]
[  0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   1.00000000e+00   0.00000000e+00]
[  0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   1.00000000e+00]
[ -0.00000000e+00   4.90500000e+01  -0.00000000e+00   4.62592927e-15
-1.15648232e-15  -3.61400724e-17  -0.00000000e+00  -0.00000000e+00
-0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]
[ -0.00000000e+00   4.90500000e+02  -1.96200000e+02  -0.00000000e+00
-0.00000000e+00   5.42101086e-16  -0.00000000e+00  -0.00000000e+00
-0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]
[ -0.00000000e+00  -2.45250000e+02   3.92400000e+02  -1.47150000e+02
-0.00000000e+00  -2.03287907e-15  -0.00000000e+00  -0.00000000e+00
-0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]
[  0.00000000e+00  -4.33680869e-14  -1.96200000e+02   2.94300000e+02
-9.81000000e+01   7.22801448e-15   0.00000000e+00   0.00000000e+00
-0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]
[  0.00000000e+00   2.16840434e-14   0.00000000e+00  -1.47150000e+02
1.96200000e+02  -4.90500000e+01   0.00000000e+00   0.00000000e+00
-0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]
[ -0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00
-9.81000000e+01   9.81000000e+01  -0.00000000e+00  -0.00000000e+00
-0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]]




In [39]:

print(B)




[[  0.00000000e+00]
[  0.00000000e+00]
[  0.00000000e+00]
[  0.00000000e+00]
[  0.00000000e+00]
[  0.00000000e+00]
[  5.00000000e+02]
[  2.50000000e+03]
[ -1.73472348e-13]
[  3.46944695e-13]
[ -1.73472348e-13]
[ -0.00000000e+00]]



Now that we have a linear system, the python-control package can be used to design an optimal controller for the system.



In [40]:

import control
from numpy.linalg import matrix_rank



First we can check to see if the system is, in fact, controllable. The rank of the controllability matrix must be equal to the number of rows in $A$, but the matrix_rank algorithm is numerically ill conditioned and for certain values of $n$ this will fail, as seen below for $n=5$. Nevertheless, the system is controllable, no matter the number of links.



In [41]:

matrix_rank(control.ctrb(A, B)) == A.shape[0]




Out[41]:

False



So now we can compute the optimal gains with a linear quadratic regulator. I chose identity matrices for the weightings for simplicity.



In [42]:

K, X, E = control.lqr(A, B, np.ones(A.shape), 1);



The gains can now be used to define the required input during simulation to stabilize the system. The input $r$ is simply the gain vector multiplied by the error in the state vector from the equilibrium point, $r(t)=K(x_{eq} - x(t))$.



In [43]:

def right_hand_side(x, t, args):
"""Returns the derivatives of the states.

Parameters
----------
x : ndarray, shape(2 * (n + 1))
The current state vector.
t : float
The current time.
args : ndarray
The constants.

Returns
-------
dx : ndarray, shape(2 * (n + 1))
The derivative of the state.

"""
r = np.dot(K, equilibrium_point - x)    # The controller
arguments = np.hstack((x, r, args))     # States, input, and parameters
dx = np.array(solve(M_func(*arguments), # Solving for the derivatives
f_func(*arguments))).T[0]

return dx



Now we can simulate and animate the system to see if the controller works.



In [44]:

x0 = np.hstack((0,
np.pi / 2 * np.ones(len(q) - 1),
1 * np.ones(len(u))))
t = np.linspace(0.0, 10.0, num=500)
x = odeint(right_hand_side, x0, t, args=(parameter_vals,))



The plots show that we seem to have a stable system.



In [45]:

lines = plt.plot(t, x[:, :x.shape[1] / 2])
lab = plt.xlabel('Time [sec]')
leg = plt.legend(dynamic[:x.shape[1] / 2])







In [46]:

lines = plt.plot(t, x[:, x.shape[1] / 2:])
lab = plt.xlabel('Time [sec]')
leg = plt.legend(dynamic[x.shape[1] / 2:])







In [47]:

animate_pendulum(t, x, arm_length, filename="closed-loop.mp4")







In [48]:

from IPython.display import HTML
html = \
"""
<video width="640" height="480" controls>
<source src="closed-loop.mp4" type="video/mp4">
</video>
"""
HTML(html)




Out[48]:



The video clearly shows that the controller can balance all $n$ of the pendulum links. The weightings in the lqr design can be tweaked to give different performance if needed.

This example shows that the free and open source scientific Python tools for dynamics are easily comparable in ability and quality to a commercial package such as Mathematica.

The IPython notebook for this example can be downloaded from https://github.com/pydy/pydy/tree/master/examples/npendulum. You can try out different $n$ values. I've gotten the equations of motion to compute for an open loop simulation of 10 links. My computer ran out of memory when I tried to compute for $n=50$. The controller weightings and initial conditions will probably have to be adjusted for better performance for $n>5$, but it should work.



In [49]:

%install_ext http://raw.github.com/jrjohansson/version_information/master/version_information.py
%version_information numpy, sympy, scipy, matplotlib, control




Installed version_information.py. To use it, type:

Out[49]:

SoftwareVersionPython2.7.10 64bit [GCC 4.4.7 20120313 (Red Hat 4.4.7-1)]IPython3.1.0OSLinux 3.13.0 54 generic x86_64 with debian jessie sidnumpy1.9.2sympy0.7.7.devscipy0.15.1matplotlib1.4.3control0.7.0Thu Jun 11 17:15:19 2015 PDT