Re: Great Explanation, Sam...
OK, so let us say that my watch is held at a distance h from the ground, that this ground, depending on its nature, offers a cushion the thickness of which is x, before the fall of the watch comes to a stop.
The KE of the watch when it hits the floor equals its potential energy when I drop it.
This energy equals the work of a stopping force F applied on the watch while it crosses the cushion.
That gives m g h = F x.
m g h = m gamma x
h =(gamma/g) x
Gamma is the max acceleration that the watch may sustain, 5000g.
That gives h = 5000 x, so it confirms that the softer the ground is (the thicker the cushion), the higher I may drop my watch, and the ratio of the two is roughly 5000. So, if I drop my watch above a sponge that would squeeze a whole 10mm at the impact, I can drop it from 50m high(???). If I drop it above concrete or a tile floor that would squeeze only, say, 0.1 mm at the impact, I can only drop it from 5m before it is broken(???).
It is not easy to calculate. It is quite safe to neglect the friction of air while the watch is airborne, but what happens at the impact site needs many parameters to be properly defined, like the damper effect of the floor and its elasticity (because part of the energy at impact will be used to make the watch bounce some times). That revives some souvenirs of old, full of nasty things called differential equations that I do not (repeat: do not) want to get back into!!!
5 meters above a concrete ground? Sounds plausible, I think the movement may well survive that, but logically something has to break: the crystal, the bezel, the crown shaft... It also depends onto which part of the watch it falls.
I need volunteers for confirming my results... anyone? ;-)
