Control & mechanics · Force limits

Cart-Pole Balance

Can a cart's motor catch a falling pole and then bring the cart back to its start? Yes, if the release tilt is small and the motor can push hard enough. To catch a pole leaning right, the cart must first accelerate right and get back under it. The push also depends on the cart's position and speed, so once the pole is upright the cart is steered back to its start. If the push needed is larger than the motor's limit, the pole falls.

Tilt a three-dimensional cart-pole and let it go. Watch the feedback controller catch the pole and bring the cart back, and see balance fail when the motor cannot push as hard as the controller asks.

Tilt from upright $0$

Rod angular speed $0$

Delivered cart force $0$

Requested cart force $0$

Cart position $0$

System horizontal momentum $0$

Signed motor work $0$

Maximum sampled work-balance residual $0$

Release & actuator

Catch the fall, then find where control fails

Initial tilt

\theta_0

8.00\,{}^\circ

Initial rod angular speed

\omega_0

0.00\,\mathrm{rad/s}

Motor off or feedback on

b_{\rm on}

1.00\,

Nominal response rate

\lambda

2.00\,\mathrm{s^{-1}}

Force limit

F_{\max}

12.00\,\mathrm N

Pivot-to-bob length

l

1.00\,\mathrm m

Integration: standard or fine

b_{\rm fine}

0.00\,

A positive tilt leans right of vertical. Drag the gold handle below to set the starting tilt, or use the sliders with the keyboard. Playback runs at one quarter of physical speed so the response is easy to follow.

The cart has a mass of one kilogram and the point bob a quarter kilogram. The drawn sizes of the cart and bob and the rod thickness are schematic. The rail is an ideal guide with no ends; the ends drawn on screen are not stops.

Full-state linear feedback puts all four poles of the linearized system at minus the nominal response rate. The force is clipped at the chosen limit. Control acts only within sixty degrees of upright, and stability near upright does not guarantee recovery from every start.

Each run covers eight physical seconds with fourth-order Runge–Kutta in 8000 or 16000 steps, plus cubic Hermite interpolation. The signed work is integrated with the same dynamics, and the residual shown checks the stored step states. Saturation and controller switching make the motion less smooth. To judge trajectory accuracy, compare with the fine integration.

The pivots are ideal and frictionless, the motion is held to a plane, and there are no collision surfaces. A rod that falls may swing through the schematic stand. Sensor delay, structural flex, electrical losses, a regenerative battery model, hardware safety guarantees and a swing-up controller are not included. Fast rotor markings can alias during real-time playback, but the angle and exported data keep physical time.

The bob is an ideal point mass on a massless rigid rod, and the rail is frictionless and endless. The controller is local state feedback with a capped force. Swing-up, impacts, sensor noise and the motor's electrical behavior are not modeled.

Start with the essentials

Focus question: Why does a rightward lean first require a rightward push?
One-sentence intuition: Accelerating the base sideways rotates the pole, because the pole is pinned to the cart. The controller also watches the cart's position and velocity, so the cart comes back to where it started.

Core mathematical model

Nonlinear cart-pole dynamics

(M+m)\ddot x+ml\cos\theta\,\ddot\theta-ml\sin\theta\,\dot\theta^2=F,\qquad l\ddot\theta+\cos\theta\,\ddot x=g\sin\theta

Tilt is measured toward the right from upright. The rod is massless and the bob is a point mass.

Full-state feedback near upright

F_d=k_xx+k_v\dot x+k_\theta\theta+k_\omega\dot\theta,\qquad F=\operatorname{clip}(F_d,-F_{\max},F_{\max})

The gains are recalculated whenever you change the length or the response rate. They put all four poles of the linearized, unsaturated system at minus the response rate. Every gain sign follows the angle convention above.

Mechanical energy and motor work

E=\tfrac12(M+m)\dot x^2+ml\dot x\dot\theta\cos\theta+\tfrac12ml^2\dot\theta^2+mgl(\cos\theta-1),\qquad\dot E=F\dot x

The cross term that couples cart and pole cannot be dropped. Potential energy is zero at upright. Only the change in energy is compared with the motor's work, so the choice of zero does not matter.

Common difficulties

Feedback has limits

Typical misconceptionOnce feedback is on, the system will recover from any release.

Better mental modelCompare the limit presets. These controllers balance only near upright and have limited drive strength. They are not swing-up controllers.

A balanced pole still needs the cart controlled

Typical misconceptionFeeding back the angle alone will bring the cart back to its start.

Better mental modelThe control force depends on the cart's position and velocity as well as the pole's tilt and angular velocity.

Run the experiment

01
Tilt and release
Drag the gold handle to set the initial tilt, then press play.
What to observe: Watch the rod, and watch how hard and in which direction the controller drives the motor.
02
Find the motor's limit
Choose Too little force, then raise the force limit.
What to observe: Look for the moment when the motor stops delivering what the controller asks for.
03
Compare energy with motor work
Go to the end of the playback and export the record. Then compare standard and fine integration.
What to observe: The change in mechanical energy matches the work done by the motor, which is negative whenever the motor brakes.

Primary theory: MIT · Cart-Pole · ETH Zürich · 1D reaction-wheel model, Section II