# Is there any robot stability control equation that describes the relation between velocity of the robot and its orientation?

Let’s assume at time t a moving robot (e.g. PIONEER 3-DX) changes it’s steering angle by $25^{\circ}$. Obviously, in order to maintain stability and avoid overturning, the robot must reduce its velocity between the time interval $t$ and $t+1$. But the question is by how much?

That's not obvious. If I'm in a tank, going 0.5 km/h, I don't need to slow down at all. If I'm in a bobsled going 100km/h and the track banks, I don't need to slow down at all.

When you steer, you begin to move around a circle with a particular radius of curvature. This means you also begin to experience centrifugal force.

$$F_c = mv^2/r$$

where $F_c$ is the centrifugal force, $m$ is the mass, $v$ is the speed, and $r$ is the radius of curvature.

Now, you have a force. It acts through the center of mass of the vehicle. If the center of mass is at some height $h$ above the surface, then you get a torque in addition to the force, where:

$$\tau_c = F_c h$$

where $\tau_c$ is the torque on the vehicle caused by the centrifugal force and $h$ is the height above the road.

Now, the centrifugal force is free to increase until one of two things happens:

1. The centrifugal force exceeds the static force between the tires and the road. Once this happens the tires "break loose" and the vehicle begins to "drift".
2. The torque exceeds the torque required to rotate the vehicle. Once this happens the vehicle rolls over.

There was a bunch of news coverage maybe 15 years ago when SUVs were first introduced in the US where the vehicles where rolling over a lot. Their center of mass was too high, meaning that the torque on the vehicle was created much earlier (at lower speeds) than drivers were used to.

So, you don't need to slow down at all until the tire's grip is overcome and/or the vehicle begins to roll.