From: Franc Zabkar on 29 Jan 2010 23:33 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 nonoperating 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 halfsine 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 30 Jan 2010 08:50 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 nonoperating 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 > halfsine 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 30 Jan 2010 12:58 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 nonoperating 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 > halfsine 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 30 Jan 2010 19:00 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 30 Jan 2010 19:16 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 nonoperating 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 > halfsine 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 highspeed 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 Gforces are small, but the destruction can be massive. Wear eye protection and maybe gloves.

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