> This whole thing reminds me of the conundrum, if that's the word, of the bird in the cage. If a bird is in an airtight cage attached to a scale and it starts to fly, will the scale read the same or less than before? I believe the answer is no as the bird still exerts downward pressure on the air below which in turn affects the scales.
> Am I right in thinking this?

Erm. As long as the cage is *airtight*, I believe it doesn't matter whether the bird is on the bottom of the cage, or hovering in the middle without touching any sides, or dangling upside-down from the ceiling pining for the fjords. The cage, and the bird and the air contained inside the cage are one unit; one unit whose mass does not change. It is a closed system. Refer to the Law of Conservation of Mass: if the bird flies in the middle of the cage, its wings are exerting a force (air pressure) which vectorially averages downwards to keep itself aloft. (Of course, if the bird really, really exerts itself until it reaches exhaustion and runs out of air, it will convert a large chunk of its bodily mass into heat, but I'm ignoring that part.) Giving the parameters of the stated question, the only reason why the weighing scale *would* register rapid oscillations in weight is due to the bird's movements inside the cage. The bird is changing its momenta and the inertia of the cage, but not the total mass hanging from the scale. That is, assuming the scale is sensitive enough to detect minor mass changes.

On the other hand, a cage that is *not* airtight, such as a wire-frame birdcage, is an open system. I see no reason why the hanging scale wouldn't read noticably less whenever the bird achieves lift-off inside the cage. The forces exerted by the bird's wings are not contained inside the framework, because a bird inside an open-air cage together do not act as one unit. Whew.