Originally posted by: TallBill
350 G's is like throwing the hard drive out the window from a bullet train into the front of another bullet train.
Originally posted by: Idontcare
A rule of thumb you can use for estimating g-forces is to ratio the distances involved if the velocity of the object was gained from gravity (dropped) in the first place and air-resistance/terminal velocity is not a significant factor in the situation.
Thusly, you state your "G budget" is 350. This means the h-drive can "fall" 350 times the distance you intend it to stop within.
For example, if you intend to stop the drive over the span of one inch (say falling into jello >1" thick but the drive stops hypothetically after plowing thru just one inch of jello) then you are allowed to drop the drive from a distance 350 times the stopping distance...i.e. 350 inches (assuming jello can indeed stop the drive in one inch, if it cannot then subsitute something else which can).
Or for a practical answer to your questions - if the 3.5" tall drive falls over from on-edge to land flat then the drive can come to a stop in as little as 3.5"/350 = 0.01" without exceeding the 350G budget. In other words it must stop over the distance of 0.01" or more but no less than 0.01" otherwise it will experience more G's.
The drive is likely to flex more than this on impact so it is quite likely to be ok. 0.01" = 0.254 mm = 254 micron
For the 3.5ft question...the drive's components have a budget of 3.5 ft/350 = 0.01 ft = 0.12 inches ~= 0.3 cm. In other words provided the hard-drive comes to a stop in no less than 0.12 inches then it will not experience more than 350 G. I am not sure if a hard-drive can safely flex 3mm...so its own inflexibility may force it to come to a stop in < 3mm thereby causing it to experience > 350G.
For 2-story question - 2-story = 35ft (hey, are you intentionally making the math easy here?)...so 35ft/350 = 0.1 ft. The hard-drive can survive a fall from a 2-story building provided it comes to a stop in no less than 0.1 foot (1.2"...see jello example above)
Originally posted by: oynaz
Originally posted by: Idontcare
A rule of thumb you can use for estimating g-forces is to ratio the distances involved if the velocity of the object was gained from gravity (dropped) in the first place and air-resistance/terminal velocity is not a significant factor in the situation.
Thusly, you state your "G budget" is 350. This means the h-drive can "fall" 350 times the distance you intend it to stop within.
For example, if you intend to stop the drive over the span of one inch (say falling into jello >1" thick but the drive stops hypothetically after plowing thru just one inch of jello) then you are allowed to drop the drive from a distance 350 times the stopping distance...i.e. 350 inches (assuming jello can indeed stop the drive in one inch, if it cannot then subsitute something else which can).
Or for a practical answer to your questions - if the 3.5" tall drive falls over from on-edge to land flat then the drive can come to a stop in as little as 3.5"/350 = 0.01" without exceeding the 350G budget. In other words it must stop over the distance of 0.01" or more but no less than 0.01" otherwise it will experience more G's.
The drive is likely to flex more than this on impact so it is quite likely to be ok. 0.01" = 0.254 mm = 254 micron
For the 3.5ft question...the drive's components have a budget of 3.5 ft/350 = 0.01 ft = 0.12 inches ~= 0.3 cm. In other words provided the hard-drive comes to a stop in no less than 0.12 inches then it will not experience more than 350 G. I am not sure if a hard-drive can safely flex 3mm...so its own inflexibility may force it to come to a stop in < 3mm thereby causing it to experience > 350G.
For 2-story question - 2-story = 35ft (hey, are you intentionally making the math easy here?)...so 35ft/350 = 0.1 ft. The hard-drive can survive a fall from a 2-story building provided it comes to a stop in no less than 0.1 foot (1.2"...see jello example above)
You are right in principle, but the math is way off. You forget all about Hooke's Law, for one ;-)
I cannot be bothered to work out the math (though it might be an interesting excercise. If any of you guys attend a physics class, your teacher might be persuaded into letting you do it).
Aside from that, my guesstimate is that you would subject a harddrive about 350 Gs by dropping it from about 2 metres onto a concrete floor. If you drop it onto a carpet or wooden floor, we are talking about a 3.5 metre drop.
The angle of impact makes a huge difference, though.
Originally posted by: superHARD
Originally posted by: TallBill
350 G's is like throwing the hard drive out the window from a bullet train into the front of another bullet train.
serious?
Originally posted by: Markbnj
Originally posted by: superHARD
Originally posted by: TallBill
350 G's is like throwing the hard drive out the window from a bullet train into the front of another bullet train.
serious?
No, not even close. See the two posts above. I know from past discussions with engineers that you can easily subject something to 100 g's by dropping it from less than your own height onto an immovable surface.
Speaking of hard disks, a long time ago I bought a new Conner 100 meg drive, for about $700 if I recall correctly. I needed to remove it and do something in the case, so I removed the clips at the end of the drive rails, and then tried to pull the cable. There was some resistance, so I pulled harder on the cable. Then I got frustrated and pulled even harder, and the cable released. The drive rails at this point turned into a launching system and the Conner flew about five feet across the bedroom, falling from a height of four feet in the process, hit the floor on a corner and bounced across to the far side of the room.
After breathing slowly and carefully for a couple of seconds I picked it up, plugged it in, and it worked fine.