Let's do some simple physics (my first attempt after 23 years)......
Assuming the watch weighs 200g or 0.2kg
Force = mass x acceleration
F= 0.2kg x 10m/s2
F= 2N
The watch fell on its face. I assume the diameter of the dial / crystal is 40mm. Hence the radius is 20mm or 0.02m. Area of the dial / crystal is therefore approx 3.1416 x 0.02m x 0.02m = 0.001256 metre squared.
Pressure exerted ==> P = F/A = 2/0.001256 = 1,592.3 Pascal = 1.5923kPa.
Now, from my memory, 1 ATM is 101.3kPa.
Can 1% of the atmosphereic pressure really crack the crystal like that? Or maybe my physics is rusty beyond help?
I think it is related to impulse force it endure rather than the force it has during its fall. Your calculation is based on during its free fall where the acceleration is the gravity. However, when it hits the group, there is a change in momentum.
Take 1m height, acceleration = 9.81 m^2 / sec and initial velocity = 0 m/s
By using one of the kinematics equation, v^2 - u^2 = 2as
v^2 - 0 = 2(9.81)(1)
v = 4.429 m/s at the last moment before impact.
momentum, p =mv = 0.2kg x 4.429m/s
=0.8859 N s
during impact, the watch was stopped almost instantaneously because there wasn't anything slowly it down and for the sake of calculation,
lets say that the impulse time is 1 x 10^-7 s (that is still not exact 0)
0.8859 N s / 0.0000001 s
Then the impulse force would be 8859000N !
let say the WR is 30 bar = 30 x 101.3x10^3 = 3039000 Pa
it means the watch can withstand force of
F = P x A = 303900 Pa x 0.001256 = 3816.984 N
So now you see the logic. If it lands on the carpet, it has more time of slowing done, to let says 0.5s , then the impulse force would be reduced to 1.7718N
Cheers