How many G ??!!

This is a discussion on How many G ??!! within the A Brief History of Cprogramming.com forums, part of the Community Boards category; While swapping a hard drive in one of my machines the other day and read the lable on it. Along ...

  1. #1
    Hamster without a wheel iain's Avatar
    Join Date
    Aug 2001
    Posts
    1,385

    How many G ??!!

    While swapping a hard drive in one of my machines the other day and read the lable on it. Along with the standard performance data and settings there was a notice.

    "Warning! Do not expose this unit to forces in excess of 75 G"
    Monday - what a way to spend a seventh of your life

  2. #2
    Mayor of Awesometown Govtcheez's Avatar
    Join Date
    Aug 2001
    Location
    MI
    Posts
    8,825
    I've seen that too...

    Stands to reason that if you and your hard drive were exposed to that much force, you'd have a lot more to worry about that keeping your HD safe.

    Does anything really produce *that* much G force? (besides, say, Jupiter)

  3. #3
    No Genius That's For Sure
    Join Date
    Aug 2001
    Location
     >' >' >' >'&>' CIS MajorIndiana CNC operator >'&>'FranceSkateboarding, Cno job7>' Swindon, UKCycling & ChessGraphic Analyst
    Posts
    127
    Where might one encounter more than that?

    I am thinking that it would have to be something that did not have people on board.

    Maybe it is like the depth limits on the back of water resistant watches. It is usually some depth that very, very few people might ever achieve.

    What kind of G's do jet fighters pull? Is there anything that experiences more than they do?

    Some kind of centrifuge type equipment maybe?

    hmmmmmmmmmm.
    He is no fool who gives up what he cannot keep in order to gain what he cannot lose.

  4. #4
    Mayor of Awesometown Govtcheez's Avatar
    Join Date
    Aug 2001
    Location
    MI
    Posts
    8,825
    Where might one encounter more than that?
    I can't think of anywhere (Jupiter's gravity is 2.5 G, btw).

    Think about it; even a small (107 lbs) person would weigh over 4 tons - this would surely kill you.

  5. #5
    Mayor of Awesometown Govtcheez's Avatar
    Join Date
    Aug 2001
    Location
    MI
    Posts
    8,825
    Maybe you found the answer, Theologian!

    I wonder what would the force be at the bottom of the Pacific, in Gs?

  6. #6
    geek SilentStrike's Avatar
    Join Date
    Aug 2001
    Location
    NJ
    Posts
    1,141
    2.5 G is 2.5 times the acceleration caused by gravity at the surface of the earth. One who weighs 100 pounds at 1 G will weight 250 pounds at 2.5 G.

    IIRC, jet fighters approach 10 Gs at their tightest turns.
    Prove you can code in C++ or C# at TopCoder, referrer rrenaud
    Read my livejournal

  7. #7
    Mayor of Awesometown Govtcheez's Avatar
    Join Date
    Aug 2001
    Location
    MI
    Posts
    8,825
    One who weighs 100 pounds at 1 G will weight 250 pounds at 2.5 G.
    Right, I know that. I meant that at 75G someone weighing 107 lbs would weight 4 tons, not at 2.5G (unless you weren't criticizing, in which case I'll slink away)

  8. #8
    geek SilentStrike's Avatar
    Join Date
    Aug 2001
    Location
    NJ
    Posts
    1,141
    Oh.. misread the post. Thought you meant they would weigh > 4 tons at 2.5 Gs.
    Prove you can code in C++ or C# at TopCoder, referrer rrenaud
    Read my livejournal

  9. #9
    Has a Masters in B.S.
    Join Date
    Aug 2001
    Posts
    2,267
    >IIRC, jet fighters approach 10 Gs at their tightest turns.

    New Russian and US fighters pull aroung 12+ G's

    75 G's hmm........................... why is that even on there?

  10. #10
    No Genius That's For Sure
    Join Date
    Aug 2001
    Location
     >' >' >' >'&>' CIS MajorIndiana CNC operator >'&>'FranceSkateboarding, Cno job7>' Swindon, UKCycling & ChessGraphic Analyst
    Posts
    127
    Originally posted by Govtcheez
    Maybe you found the answer, Theologian!

    I wonder what would the force be at the bottom of the Pacific, in Gs?
    You really wouldn't look at it that way.

    The concern at the bottom of the pacific would be one involving pressure whereas G's are associated with acceleration. I was just thinking that this would be similar in that the limit is theoretical rather than tested.

    Based on what's been posted so far- it looks like the 75 G limit is one that has been calculated, not tested.

    I've been trying to think of what pulls Gs. Fighters come to mind because they accelerate hard and I worked on them. Rockets pull some but I don't think anything like 75.

    The only other thing I can think of is small machinery rotating at a high speed. Research equipment or something.

    I've never seen this on a HD before but I think that it is really interesting. May have to do some more investigating- contact some manufacturers.
    He is no fool who gives up what he cannot keep in order to gain what he cannot lose.

  11. #11
    Unregistered
    Guest
    If you drop the harddrive in the concrete floor it will go from fairly high speed to 0 in a couple of mm. That would produce a very high G force. Why they choose to frase it in G's I don't get. How about a label saying " Don't drop the hard drive on the floor"
    ~Barjor

  12. #12
    Barjor
    Guest
    Ok I am back after some work with the old pen and paper (yeap I still have them) If you drop the hard drive from 1 m. I then assumed it deaccelerate in 2 mm (landing flat on the back) That would produce 500G. To make it under 75 G it would have to stop on 13 mm (1/2")

  13. #13
    geek SilentStrike's Avatar
    Join Date
    Aug 2001
    Location
    NJ
    Posts
    1,141
    What do you mean by "deaccelerate in 2 mm? "

    Shouldn't you be measuring the deacceleation in time rather than distance?
    Prove you can code in C++ or C# at TopCoder, referrer rrenaud
    Read my livejournal

  14. #14
    Unregistered
    Guest
    Well deacceleration is measured in m/s*s (don't know hot to make a square thingy in here) My assumption was that it take the hard drive 2 mm to come to a complete stop due to deflection in the floor and hard drive. That can be recalculated to the time it take for the hard drive to stop.
    ~Barjor

Popular pages Recent additions subscribe to a feed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21