TL;DR:
Static balance keeps the center of mass over the support polygon at every instant, so the robot is stable if it freezes.
Dynamic balance accepts that the center of mass sometimes leaves the polygon and uses motion and fast control to catch the fall.
A fall develops on a time scale of about √(h / g), which is a fraction of a second for a small robot, so the control loop has to be much faster.
A legged robot can balance in two quite different ways. One is like a table: it stands because of where its feet are, and it would stay up if you cut the power to the controller. The other is like a person riding a bicycle: it stays up only because it keeps acting. Knowing which kind your robot needs shapes the whole design.
Static balance
A robot is statically stable when the ground point under its center of mass (COM) lies inside its support polygon (see center of mass and why robots tip over). Its static stability margin is the shortest distance from that point to the polygon's edge. A bigger margin means more room for error.
To walk statically, the robot shifts its COM into the support polygon before it lifts a foot, then steps, then shifts again. At each instant it could stop and stay up. That makes statically stable walking forgiving, slow and easy to program.
Dynamic balance
When a robot runs, trots, hops or walks on two legs, there are moments when the COM is not over the polygon, or the polygon is just a line or a point. The robot is falling, and it recovers by moving a foot to a new place before the fall completes. This is dynamic balance.
A useful model is the inverted pendulum: a point mass (the COM) on top of a massless leg that pivots at the foot. If the pendulum leans, gravity pulls it further over. The lean grows exponentially, with a time constant that depends on the height of the COM h and gravity g:
τ = √(h / g)
The lean roughly multiplies by e (about 2.7) every τ seconds. This is the standard result for the linear inverted pendulum model.
Worked example: standing still versus walking
A small legged robot has its COM at h = 0.30 m and feet at the corners of a rectangle 0.40 m long and 0.20 m wide.
Standing. All four feet are down. The COM is in the center, so the closest edges are the sides, 0.10 m away. The static margin is 0.10 m.
Walking statically. With one foot lifted, the polygon becomes a triangle. We computed a similar case in how a quadruped robot keeps its balance: the COM must shift to give a margin of a few centimeters, say 0.045 m. That is less than half the standing margin of 0.10 m. If a push or an error moves the COM by 0.045 m toward the edge, the robot starts to tip.
Falling time. For the inverted pendulum, τ = √(0.30 / 9.8) = √0.03061 = 0.175 s.
Suppose a push makes the COM drift 0.045 m past the edge. The time constant tells you that the lean will grow by a factor of e in about 0.175 s, and by a factor of about 20 in 3τ = 0.525 s. So the robot has maybe a few tenths of a second to place a foot and recover.
To react in time, the controller must sense and act much faster than τ. A loop that runs 10 times per τ is 10 / 0.175 = 57 Hz, and in practice robots run balance loops at hundreds of Hz or more, because sensor filtering and motor response eat into the available time. A control loop running at 5 Hz would be hopelessly slow: 1 / 5 = 0.2 s between updates is longer than τ.
Walking dynamically. In a trot, the robot is supported by two diagonal feet at a time. Stability is no longer about the margin at one instant. Instead, the controller places the next foot where the COM is heading. The margin is replaced by a rule about where to step. (A number often used for this is the capture point, the place where the robot would have to put its foot to come to a stop. For the inverted pendulum, it is x_cp = x + v × √(h / g), so at a speed of 0.5 m/s, 0.5 × 0.175 = 0.0875 m ahead of the COM.)
Where it goes wrong
Sensing delay. Filtering the IMU adds delay (see sensor noise and how to smooth it). Too much and the fall is already under way.
Motor limits. The legs need enough torque and speed to push the body back. A small servo at a high gear ratio may be too slow to catch a fall.
Foot contact detection. The controller must know when the foot touches down. A late or false contact message causes a stumble.
Terrain. Dynamic balance on soft or uneven ground needs a tolerance for foot slip.
Choosing the wrong kind. A robot that is built for dynamic balance but has a slow controller will fall. A statically stable design is safer for a first project.
Safety: Dynamic balance means a robot that can fall without warning when a sensor or the code fails. Use a harness, a hanging stand or a ceiling line for early tests, keep hands and faces away from the legs, and make sure the motors are cut if the controller stops responding. Check batteries and wiring before each run, since a loose wire can cause a fall.
How real robots use both
Many commercial legged robots stand statically and walk dynamically. Hobby hexapods, with six legs, can walk almost entirely statically because at least three feet are always down. Two-legged humanoids need dynamic balance from the start, since a polygon of two feet is small and the COM is high. Flying robots are the ultimate unstable system, which is why how a quadcopter flies depends on very fast stabilization.
Try it yourself
Balance a broom handle on the palm of your hand and note how you move your hand. Estimate how long it takes to fall if you stop moving. Then try with a shorter stick: it falls faster, as the √(h / g) formula predicts. Calculate τ for a 1 m stick whose center of mass is at 0.5 m: √(0.5 / 9.8) = 0.226 s, then for a 0.3 m stick whose center is at 0.15 m: √(0.15 / 9.8) = 0.124 s. The shorter one is harder to balance.
Frequently asked questions
Is a statically stable robot always better?
It is safer and simpler, but slower and limited in the terrain it can cross. Dynamic balance gives speed and agility at the cost of control complexity.
What is the inverted pendulum used for?
It is a simple model of a balancing body: a mass on top of a leg. It predicts how quickly a fall grows and where to place a foot to stop it.
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