[Zimmerman]

I was moving things around between my PC and my Mac and in the process of fidding with my pc, my 80gb Maxtor drive got pushed off the desk onto a lenolium floor. I am running a Norton surface scan right now. Can I ever trust this drive again, or is it time to break out the RMA forms?

[engage1000]

Ouch! I'd want to be very leery. What I would do it put it back in the computer and do a surface scan. It will take many hours, but it will check every single part of that drive physically. If it comes out okay... then you are good to go. If not, well you know.

That's a tough call. It more than likely is okay, but check it all out anyways. If it has come up with errors... RMA the sucker.

[Zimmerman]

Originally posted by engage1000:
Ouch! I'd want to be very leery. What I would do it put it back in the computer and do a surface scan. It will take many hours, but it will check every single part of that drive physically. If it comes out okay... then you are good to go. If not, well you know.

That's a tough call. It more than likely is okay, but check it all out anyways. If it has come up with errors... RMA the sucker.

Thanks. I dunno if it'll take HOURS. Its 2/3 done after 45 minutes.

[engage1000]

You're welcome Zim.
I was just looking at Maxtor's datasheet for their DiamondMax Plus 9 lineup... apparently those drives can withstand a non-operating shock of 300 G's for 2 milliseconds. Can any mathematicians or physicists here tell us if an object weighing 1.27 pounds or 630 grams can withstand a fall of 3 feet or so at the speed induced by gravity on such a device with this weight?

[Zimmerman]

Originally posted by engage1000:
You're welcome Zim.
I was just looking at Maxtor's datasheet for their DiamondMax Plus 9 lineup... apparently those drives can withstand a non-operating shock of 300 G's for 2 milliseconds. Can any mathematicians or physicists here tell us if an object weighing 1.27 pounds or 630 grams can withstand a fall of 3 feet or so at the speed induced by gravity on such a device with this weight?

Wow. Thats acutally pretty tough. I have an 8gb drive that is rated for 79.2gs.

We do know the appropriate information. What we need to know is the accleration function and plug in the appropriate values for the variables and take the derivateve to find the instantaneous velocity for f(x)=0. then we multiply the veloctiy by mass to find total intertia and since we know the drive stopped pretty much instantaneous (about 2ms), we will know whether or not it exceeded its own weight by 300x or less.

[engage1000]

Originally posted by Zimmerman:
Wow. Thats acutally pretty tough. I have an 8gb drive that is rated for 79.2gs.

We do know the appropriate information. What we need to know is the accleration function and plug in the appropriate values for the variables and take the derivateve to find the instantaneous velocity for f(x)=0. then we multiply the veloctiy by mass to find total intertia and since we know the drive stopped pretty much instantaneous (about 2ms), we will know whether or not it exceeded its own weight by 300x or less.

How's the surface scan going?

[Zimmerman]

Originally posted by Zimmerman:
Wow. Thats acutally pretty tough. I have an 8gb drive that is rated for 79.2gs.

We do know the appropriate information. What we need to know is the accleration function and plug in the appropriate values for the variables and take the derivateve to find the instantaneous velocity for f(x)=0. then we multiply the veloctiy by mass to find total intertia and since we know the drive stopped pretty much instantaneous (about 2ms), we will know whether or not it exceeded its own weight by 300x or less.

Ok, I asked a guy down the hall (dorm) and he said the accelration function of the pull of Earth's gravity is f(s)=9.8s^2
where s=seconds, and 9.8=meters.

So what we need to do is convert 3ft into meters (within 100ths of a meter) and figure out how many seconds it would take to fall that distance. Then we plug that value into the function above and we arrive at the approximate instantaneous velocity when the drive hit the floor.

[Zimmerman]

Originally posted by engage1000:
How's the surface scan going?

It finished about 10 minutes ago; I got my Expert Rifleman qualification in AAO while it scanned; I think the stress calmed my hands. Anyway, the drive works and is fine !!!!! No errors.

[Zimmerman]

3ft in meters is about .9144. I don't know how to apply it to the f(x)=9.8s^2 formula. We can't set it equal to it since f(x) is given as a velocity, and a distance does not equal a velocity.

[rjenkinson]

f=ma

the force of the impact (in newtons) is equal to the mass (in grams) multiplied by the acceleration.

remember to convert pounds to kilograms. one kilogram is equal to 2.2 pounds.

-r.

[Zimmerman]

Originally posted by rjenkinson:
f=ma

the force of the impact (in newtons) is equal to the mass (in grams) multiplied by the acceleration.

remember to convert pounds to kilograms. one pound is equal to 2.2 kg.

-r.

So would the calculation go:

0.9144=9.8s^2
---->
s ~= .30546 (it takes under a third of a second for the hdd to fall 3 ft.

Now we need to take a limit (derivative) of this equation

lim 9.8(s)^2
s->0.30546

which of course, using the difference quotient

lim [f(h) - f(.030546)]/h^2
h->0

but I'm not sure how to solve that...

btw 1.72lbs = 2.794kg

[Mac Zealot]

as long as the drive is not spinning, it's cool.

I've had more than my fair share of hard drive drops. Once a 3gb quantum I had for my pc was sitting on my old table, on a pile of stuff, somehow I was out, the window was open, no screen, and the hard drive fell out!

When I got home I was shocked to see the thing on path outside the house, I took it in and checked it, perfectly fine, though it was dinged and the pins were messed up (took a bit of work with a screwdriver)

had a laptop fall from 7ft on a shelf too.

[Ver de Terre]

Originally posted by Zimmerman:
So would the calculation go:

0.9144=9.8s^2
---->
s ~= .30546 (it takes under a third of a second for the hdd to fall 3 ft.

Now we need to take a limit (derivative) of this equation

lim 9.8(s)^2
s->0.30546

which of course, using the difference quotient

lim [f(h) - f(.030546)]/h^2
h->0

but I'm not sure how to solve that...

btw 1.72lbs = 2.794kg

d = v0 * t + 1/2 * a * t ^ 2 ->
t = (2 * d / a) ^ 0.5
because v0 = 0m/s

So,
t = (2 * 0.91m / 9.8m/s^2) ^ 1/2 = 0.43s

And also, 1.72lbs = (1.72 / 2.2)kg = 0.78kg

Glad to hear the thing works

[rjenkinson]

Originally posted by Zimmerman:
So would the calculation go:

0.9144=9.8s^2...

actually, there is no time in the equation per se. that 9.8 seconds squared is really a constant squared so you don't need time to calculated the force of the impact. it should look something like this:

f=ma
f=780 x 9.8 x 9.8
f=74911 N

-r.

[Ver de Terre]

Originally posted by rjenkinson:
actually, there is no time in the equation per se. that 9.8 seconds squared is really a constant squared so you don't need time to calculated the force of the impact. it should look something like this:

f=ma
f=780 x 9.8 x 9.8
f=74911 N

-r.

In that equation, m is in kilograms, not grams, so F = 0.78kg * 9.8 m/s^2 = 7.6N. But that doesn't represent the actual damage, which would require an energy calculation.

If the force were 75kN, I think the entire room would look a bit different.

[rjenkinson]

Originally posted by Ver de Terre:
If the force were 75kN, I think the entire room would look a bit different.

whoops! i thought i was reading the "can the room survive a hard drive falling 2.5-3ft fall?" thread.

-r.

[SOLIDAge]

i dropped my firewire drive off my desk, atleast 3 feet
makes a little noise ahahah but works fine and has for awhile

[wataru]

We can't know the force imparted on the drive unless we know either how long the collision took or how high the drive bounced or how fast it was going immediately after the collision.

Yes, F=ma, but that only gives the force of the earth on the drive (and the drive on the earth). The pertinent equation in this instance is F=∆p/∆t, (force is equal to the change in momentum divided by the change in time). Unless we can tell how much the momentum changed (and how quickly), we don't know the force imparted on it.

[AlbertWu]

this takes some math. if you know the velocity the INSTANT before it hit the ground, and how long it took to stop, you could find the "a" part of the f=ma formula, plug in the mass in kilograms, and convert the resulting "f" into Gs by using 9.8

figuring out the instantaneous velocity before the hit is the hard part.

[swsteckly]

I have taken my fair share of HDD's apart (failed ones- for fun). The Maxtor DiamondMax Plus 9 is by far the sturdiest looking of them all. It uses a six-bolt hub, and all of the arms are heavily braced.

I have had a Diamond Max 16 (5400RPM, 2MB) 160GB on my desk for a few years(Maxtor enclousure), for general data storage/backup.

I have also had a DiamondMax Plus 9 (7200RPM, 8MB) 120GB for a year (LaCie enclousure), for DV import (where speed matters).

Both have worked flawlessly since I got them. I have had many IBM and Western Digital faillures over the years, but these appear to be much more reliable (I know that technology has improved, but still...).

[ARENA]

Originally posted by Mac Zealot:
as long as the drive is not spinning, it's cool.

I've had more than my fair share of hard drive drops. Once a 3gb quantum I had for my pc was sitting on my old table, on a pile of stuff, somehow I was out, the window was open, no screen, and the hard drive fell out!

When I got home I was shocked to see the thing on path outside the house, I took it in and checked it, perfectly fine, though it was dinged and the pins were messed up (took a bit of work with a screwdriver)

had a laptop fall from 7ft on a shelf too.

I see you take care of your stuff.

[wataru]

Originally posted by AlbertWu:
this takes some math. if you know the velocity the INSTANT before it hit the ground, and how long it took to stop, you could find the "a" part of the f=ma formula, plug in the mass in kilograms, and convert the resulting "f" into Gs by using 9.8

figuring out the instantaneous velocity before the hit is the hard part.

No, it's not hard at all. It had a potential energy of mgh before it fell, and all of that energy was converted into kinetic energy by the time it hit the floor.

mgh=.5mv^2
v=(2gh)^.5

But we also need to know fast it was going AFTER the collision (again, F=∆p/∆t). This we could find by knowing how high it bounced.

Using F=ma is not the way to go about solving this problem.

Edit: Ok, let's assume it didn't bounce, it fell from 1m, its mass is .5kg, and it took 0.002s to stop. Using g=10m/s^2

v=(20)^.5≈4.5m/s
F=∆p/∆t=(.5*4.5-.5*0)/.002=1125N

Its weight due to gravity is about 5N, so assuming all of these variables, it would have received a force 225 times its normal weight.

I think the assumed ∆t is the most flaky of these numbers, and it has the biggest effect on the outcome, so this number is probably wrong.

[Zimmerman]

Originally posted by wataru:
No, it's not hard at all. It had a potential energy of mgh before it fell, and all of that energy was converted into kinetic energy by the time it hit the floor.

mgh=.5mv^2
v=(2gh)^.5

But we also need to know fast it was going AFTER the collision (again, F=∆p/∆t). This we could find by knowing how high it bounced.

Using F=ma is not the way to go about solving this problem.

Edit: Ok, let's assume it didn't bounce, it fell from 1m, its mass is .5kg, and it took 0.002s to stop. Using g=10m/s^2

v=(20)^.5≈4.5m/s
F=∆p/∆t=(.5*4.5-.5*0)/.002=1125N

Its weight due to gravity is about 5N, so assuming all of these variables, it would have received a force 225 times its normal weight.

I think the assumed ∆t is the most flaky of these numbers, and it has the biggest effect on the outcome, so this number is probably wrong.

ack! all those sigma-hat, delta-hat... ack! I don't wanna hear about elasticities any more!!! Lol.

I'm an Economics student, and this is what I look at. I'm only in the first of two Calculus courses I'll wind up taking, so my calc skills are appreciably weak.

[CharlesS]

You might want to tweak Norton's settings so that the surface scan does Read, Write, and Verify instead of just Read.

This will make it take forever, but this always used to turn up bad sectors in floppies where the read scan would report no errors, back in the day...

[wataru]

Originally posted by Zimmerman:
ack! all those sigma-hat, delta-hat... ack! I don't wanna hear about elasticities any more!!! Lol.

I'm an Economics student, and this is what I look at. I'm only in the first of two Calculus courses I'll wind up taking, so my calc skills are appreciably weak.

There's no calc! In this case ∆ is macroscopic so it's not derivative. ∆p is just the change in p and ∆t is just the change in t.