some 4mm, if I calculated correctly

Oh, Dragi, you knew I couldn't resist the challenge, didn't you.

Let's see... V^2 = 2 x A x D where A = 980 cm/sec/sec and D = 100 cm

That gives... 442.72 cm/sec velocity after a drop of 1 meter.

Now, let's solve it the other way:

D = V^2 / 2A, where V = 442.72 cm/sec, and A = (980 cm/sec/sec) x 500. (that's your 500 Gs deceleration)

I get (442.72 x 442.72) / (2 x 980 x 500), or 196,000 / 980,000 and once all the units cancel out that comes to .2 CM, or 2 mm.

I'm guessing you dropped the "2" in the divide by 2A step?

Anyway, 2mm of deceleration would be enough to cushion it for a 500G stop. Calculating the actual G-load of a drop onto concrete is problematic, since the only "cushion" as such is provided by the deformation of the case and it's internal parts, and of course it is the deformation of those internal parts that does the damage to the drive.

Yeah, I know -- I should get a life!

tanstaafl.