have you actually read this page?

> If the current exceeds the rated load for a prolonged period, the heat causes the bimetallic strip to bend.

> The unit trips after a delay inversely related to the magnitude of the overload; the greater the current, the faster the trip.

The heat is generated by the breaker itself, from the current. Maybe it'll trip slightly faster if the wire is very hot, but I say it's unlikely - the thermal resistance on a regular breaker is very low, and wire is not a very great heat conductor either.

And even if breaker was for some reason reacting on wire temperqture? Look at the original post. You can easily see how far the heat traveled simply by seeing where the insulation got blackened. It looks like just a few cm - nowhere close to get to the other end of the wire, which was in the breaker. The system had breaker, and it could not detect this condition.

(There are special "thermal fuses" which react on the external temperature, but they are very distinct from the regular circuit breaker, and they must be attached directly to the device emitting heat, like a motor. No one attaches those to the regular terminal strips)

>The heat is generated by the breaker itself, from the current.

Why do you think this?

>You can easily see how far the heat traveled simply by seeing where the insulation got blackened. It looks like just a few cm

Why do you think this? Copper is excellent at thermal conductivity, only beaten by silver.