How can I calculate the maximum safe drop height for a laptop rated at 1500 gs?

In summary, the conversation discusses the 1500 gs rating of a laptop and how to calculate the maximum safe drop height based on this rating. It is explained that the "g" rating refers to acceleration, and the harder the surface, the faster the deceleration when dropped. The average acceleration can be calculated using kinematic equations and the contact time between the laptop and surface. It is suggested to research "collision physics" for more accurate calculations. The speaker also mentions that laptops typically do not survive dropping onto hard surfaces, even from short heights.
  • #1
Deina
1
0
A certain laptop says it can withstand 1500 gs.

Assuming that it's dropped onto a hard surface, and ignoring air resistance, how would I calculate [STRIKE]how many times Hubby can come home late without calling[/STRIKE] the max "safe" drop height?
 
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  • #2
Welcome to PF;
That's good thinking. You basically want to know if 1500g is a lot or a little or what? The number doesn't say anything unless you understand it.

The "g" rating refers to an acceleration - or, in this case, a deceleration, in multiples of the acceleration of gravity.
The harder the surface, the faster the deceleration when you drop it.

Back of envelope, you can work out the average acceleration the laptop undergoes if you know it's initial and final velocities and the amount of time it spends decelerating.

You can get the speed just before it hits by kinematic equations.

The time you want would be the "contact time".
If we guess that the laptop and surface are hard enough for a collision (that does not break things) is very nearly elastic, then the final speed is about the same as the initial speed after the contact time is up: i.e. you expect the laptop to rebound. (This is usually a bad approximation - look up "coefficient of restitution", but it should be good enough to get the idea.)

So you can write a=2v/T (average) if T is the contact time, and v is the speed from falling height h.
The value of T depends on the details of the surface and it is something that gets measured. You can look up typical contact times for hard surfaces ... usually of the order of a few milliseconds.

(You will want to look up "collision physics")
It looks to me like you laptop is not rated for drops bigger than 1cm, but I could have misplaced a decimal point.

From experience, laptops do not long survive striking hard surfaces like husband's heads even from quite short drops like from above a door. Just saying.
 
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1. What is the formula for converting g forces to height?

The formula for converting g forces to height is h = (g * t^2)/2, where h is the height in meters, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.

2. How do you calculate the height from a given g force?

To calculate the height from a given g force, you can use the formula h = (g * t^2)/2, where h is the height in meters, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.

3. Is there a difference in the calculation for different units of g force?

Yes, there is a difference in the calculation for different units of g force. The formula for converting g forces to height remains the same, but the value of g will vary depending on the unit of measurement. For example, if g is measured in feet per second squared, the value will be 32.2 ft/s^2 instead of 9.8 m/s^2.

4. Can g forces be converted to height on any planet?

Yes, g forces can be converted to height on any planet as long as the acceleration due to gravity on that planet is known. The formula h = (g * t^2)/2 can be used, but the value of g will vary depending on the planet's gravitational pull.

5. Can g forces be converted to height in outer space?

No, g forces cannot be converted to height in outer space as there is no acceleration due to gravity. The formula h = (g * t^2)/2 requires the value of g to be present. In outer space, objects are in a state of weightlessness and do not experience the effects of gravity, so this formula would not be applicable.

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