From: Franc Zabkar on
I've never understood the shock specs of hard drives. Does anyone know
the practical meaning of the following spec?

Barracuda 7200.10 Serial ATA Product Manual:
http://www.seagate.com/staticfiles/support/disc/manuals/desktop/Barracuda%207200.10/100402371f.pdf

2.9.5.2 Nonoperating shock - 3 and 4 Disc models

"The non-operating shock level that the drive can experience without
incurring physical damage or degradation in performance when
subsequently put into operation is 300 Gs based on a nonrepetitive
half-sine shock pulse of 2 msec duration."

"All shock specifications assume that the drive is mounted securely
with the input shock applied at the drive mounting screws. Shock may
be applied in the X, Y or Z axis."

AFAICT, the numbers for a 750GB hard drive weighing 720g work out as
follows.

The impulse energy is defined by ...

E = F.dt = m.a.dt

where F = force, a = acceleration, m = mass, dt = impulse duration

So E = 0.720 kg x (300 x 9.8 m/s/s) x 0.002 sec = 4.23 Joules

If this energy is imparted by a projectile of mass M travelling at V
m/s, then ...

E = 1/2 . M . (V x V)

If M = 1 kg, then V = 2.91 m/s = 10.5 km/h

So the drive will survive a strike by a projectile weighing 1kg and
travelling at 3m/s or 10 kph???

Sounds more like a wrecking ball to me ...

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.
From: Arno on
Franc Zabkar <fzabkar(a)iinternode.on.net> wrote:
> I've never understood the shock specs of hard drives. Does anyone know
> the practical meaning of the following spec?

> Barracuda 7200.10 Serial ATA Product Manual:
> http://www.seagate.com/staticfiles/support/disc/manuals/desktop/Barracuda%207200.10/100402371f.pdf

> 2.9.5.2 Nonoperating shock - 3 and 4 Disc models

> "The non-operating shock level that the drive can experience without
> incurring physical damage or degradation in performance when
> subsequently put into operation is 300 Gs based on a nonrepetitive
> half-sine shock pulse of 2 msec duration."

> "All shock specifications assume that the drive is mounted securely
> with the input shock applied at the drive mounting screws. Shock may
> be applied in the X, Y or Z axis."

> AFAICT, the numbers for a 750GB hard drive weighing 720g work out as
> follows.

> The impulse energy is defined by ...

That is a 2ms pulse, not impulse. Thats something differently.

> E = F.dt = m.a.dt

Ok, just kinetic energy.

> where F = force, a = acceleration, m = mass, dt = impulse duration

> So E = 0.720 kg x (300 x 9.8 m/s/s) x 0.002 sec = 4.23 Joules

> If this energy is imparted by a projectile of mass M travelling at V
> m/s, then ...

> E = 1/2 . M . (V x V)

> If M = 1 kg, then V = 2.91 m/s = 10.5 km/h

> So the drive will survive a strike by a projectile weighing 1kg and
> travelling at 3m/s or 10 kph???

Wrong intepretation. The important infomation is in the
impact time. Less than 2ms and the drive may die, more and
it can be fine. If you fire a bullet at it, it will have
far less impact time. If you take it for a ride on the
motorway, going 150km/h (about 42m/s), it will not care
at all, because of very long impact (acceleration) time.,

Or if you hit it with a metal hammer it will likely die,
but the same blow with a rubber hammer may be survivable.

> Sounds more like a wrecking ball to me ...

A wrecking ball with a thin layer of rubber is fine, as
3m/s is really slow. Drop the drive from 50cm and it will
have that speed on impact.

Doing this exactly is pretty hard and requires a lot of
knowledge about meterial properties, e.g. of a surface
dropped on.

Here is a rule of thumb: Scale deaceleration length so that it
matches acceleration length times acceleration difference.
This is physically sound but does not take into account
impact time and non "clean" impact.

Say you deacellerate your disk on 1mm (hard rubber surface),
you can accelerate it for 300mm (30cm) at 1G if deaceleration
can be up to 300G. This could also be one layer of thin
foam rubber wrap.

Say you deaccelerate on 0.1 mm (metal, glass, etc.),
you can trop it about 3cm.

All very rough, but gives you an idea that 300G is not
all that much.

Arno
--
Arno Wagner, Dr. sc. techn., Dipl. Inform., CISSP -- Email: arno(a)wagner.name
GnuPG: ID: 1E25338F FP: 0C30 5782 9D93 F785 E79C 0296 797F 6B50 1E25 338F
----
Cuddly UI's are the manifestation of wishful thinking. -- Dylan Evans
From: sambo on
Franc Zabkar wrote:
> I've never understood the shock specs of hard drives. Does anyone know
> the practical meaning of the following spec?
>
> Barracuda 7200.10 Serial ATA Product Manual:
> http://www.seagate.com/staticfiles/support/disc/manuals/desktop/Barracuda%207200.10/100402371f.pdf
>
> 2.9.5.2 Nonoperating shock - 3 and 4 Disc models
>
> "The non-operating shock level that the drive can experience without
> incurring physical damage or degradation in performance when
> subsequently put into operation is 300 Gs based on a nonrepetitive
> half-sine shock pulse of 2 msec duration."
>
> "All shock specifications assume that the drive is mounted securely
> with the input shock applied at the drive mounting screws. Shock may
> be applied in the X, Y or Z axis."
>
> AFAICT, the numbers for a 750GB hard drive weighing 720g work out as
> follows.
>
> The impulse energy is defined by ...
>
> E = F.dt = m.a.dt
>
> where F = force, a = acceleration, m = mass, dt = impulse duration
>
> So E = 0.720 kg x (300 x 9.8 m/s/s) x 0.002 sec = 4.23 Joules
>
> If this energy is imparted by a projectile of mass M travelling at V
> m/s, then ...
>
> E = 1/2 . M . (V x V)
>
> If M = 1 kg, then V = 2.91 m/s = 10.5 km/h
>
> So the drive will survive a strike by a projectile weighing 1kg and
> travelling at 3m/s or 10 kph???

Nope, the drive will survive moving at that speed and hitting a fixed
object like the floor if you allow for the real weight of the system etc.

> Sounds more like a wrecking ball to me ...
>
> - Franc Zabkar


From: Franc Zabkar on
On Sat, 30 Jan 2010 15:33:19 +1100, Franc Zabkar
<fzabkar(a)iinternode.on.net> put finger to keyboard and composed:

>The impulse energy is defined by ...
>
>E = F.dt = m.a.dt
>
>where F = force, a = acceleration, m = mass, dt = impulse duration

Brain fart!

F.dt = m.a.dt = m.dv/dt.dt = m.dv

ie F.dt does not have the units of energy. It in fact represents a
change in momentum.

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.
From: mike on
Franc Zabkar wrote:
> I've never understood the shock specs of hard drives. Does anyone know
> the practical meaning of the following spec?
>
> Barracuda 7200.10 Serial ATA Product Manual:
> http://www.seagate.com/staticfiles/support/disc/manuals/desktop/Barracuda%207200.10/100402371f.pdf
>
> 2.9.5.2 Nonoperating shock - 3 and 4 Disc models
>
> "The non-operating shock level that the drive can experience without
> incurring physical damage or degradation in performance when
> subsequently put into operation is 300 Gs based on a nonrepetitive
> half-sine shock pulse of 2 msec duration."
>
> "All shock specifications assume that the drive is mounted securely
> with the input shock applied at the drive mounting screws. Shock may
> be applied in the X, Y or Z axis."
>
> AFAICT, the numbers for a 750GB hard drive weighing 720g work out as
> follows.
>
> The impulse energy is defined by ...
>
> E = F.dt = m.a.dt
>
> where F = force, a = acceleration, m = mass, dt = impulse duration
>
> So E = 0.720 kg x (300 x 9.8 m/s/s) x 0.002 sec = 4.23 Joules
>
> If this energy is imparted by a projectile of mass M travelling at V
> m/s, then ...
>
> E = 1/2 . M . (V x V)
>
> If M = 1 kg, then V = 2.91 m/s = 10.5 km/h
>
> So the drive will survive a strike by a projectile weighing 1kg and
> travelling at 3m/s or 10 kph???
>
> Sounds more like a wrecking ball to me ...
>
> - Franc Zabkar
IMHO, the operating spec is far more relevant.
Wanna have some fun?
Take an ordinary high-speed CD drive.
Plop in a cd, that you don't care about losing,
spin it up to full speed and quickly flip it
over. The G-forces are small, but the destruction
can be massive.
Wear eye protection and maybe gloves.