Preface

This analysis was initially undertaken as Omega were marketing their “Mission to Mars” Speedmaster X33with a shock-resistance rating of 10,000 g’s. I read many blog articles, reviews, and forum discussions inwhich many contemplated that it was impossible for an object to experience such a large number of ”Gs”.In many of these, there were references to the fact that fighter pilots tend to blackout before G-LOC ataround 5 g and red out during -2 to -3 g (the negative sign denotes the upward direction).

It’s all to do with the impulse of a force. An intuitive conclusion does not consider the duration and distancethe object travels during it’s deceleration from impact velocity to rest. A regular floor would be pretty hard,and as you can see, about 100-times to 1,000-times more g’s are experienced than one would typicallyexpect.

Of course, this assumes that the falling object is infinitely hard, and in practice, a falling metal object willdent; this can be called the “crumpling” distance. If this dent is sufficiently large enough, which it mostlikely will be, all the parts of the object that did not directly impact the floor will experience a muchlower deceleration, i.e., typically <1000g for a 100 g object falling through 1 m onto a granite floor.

— MichaelJanuary 2010

Comments, Criticism, and Contribution

If you have any suggestions or discover errors in this document, write to michael@mwdesilva.com tohelp make this document better. Thanks in advance for your contribution.

List of contributors:

1. Rave (RavemasterX@hotmail.co.uk): he suggested that accounting for the “crumpling” distance willshow that other parts of the object – that are not in direct contact with the point of impact – willexperience a much lower deceleration. The point of impact will crumple and thereby allow the otherareas to decelerate over a longer distance.

Michael M. W. de Silva MSc (Dist) BEng (Hons) MIEEE MIET AMIMechE(UK) MASMEemail: michael@mwdesilva.com • resume: mwdesilva.com & linkedin.mwdesilva.com

1 Impact Analysis of a Falling Object

The following assumptions are considered for the following analysis. The object weight, m, is 0.15 kg (or150 grammes), the acceleration due to gravity, g, is 9.81 ms−2, and finally the height from which it falls, s,is 1 m.

From the equations of linear motions it can be calculated as follows,

s = ut +1

2at2 (1)

The time taken for the object to fall from a height of 1 m, and it is only after this time has passed that theobject experiences ‘impact’, is calculated as follows,

t =

√2s

a=

√2

9.81= 0.45 s (2)

The impact velocity maybe obtained by equating Kinetic Energy = Potential Energy,

1

2mv2 = mgh

v =√

2gh = 4.43 ms−1

(3)

or, since,

v = a · t = 9.81

(√2

9.81

)= 4.43 ms−1 (4)

Thus, the Kinetic Energy just before impact is,

KE =1

2mv2 = 1.47 Nm (5)

The Work-Energy Principle states that, “The change in the kinetic energy of an object is equal to the network done on the object.” For a straight-line collision, the net work done is equal to the average force ofimpact times the distance traveled during the impact,

Average impact force x distance traveled = change in kinetic energy

This can be mathematically described as follows,

FAvg · d = ∆KE = KEFinal −KEInitial (6)

1

Hence,

FAvg =1.47

d=

1.47

0.01× 10−3= 147, 000 N (7)

From this, the deceleration (denoted by the negative sign),

a = −F

m= −147, 000

0.15= −980, 000 ms−2 → 980, 000

9.81= 99, 898.06 g (8)

From a = (v − u)/t, the duration of the deceleration is,

t =(v − u)

a=

4.43

99, 898.06= 0.04 ms (9)

Putting this into perspective, we can consider the power generated. It is known that KE = 1 J = 1 Nmand that 1 hp = 745.7 W . Since 1 W = 1 J/s = 1 Nm/s,

1.47(4.43

99, 898.06

) = 33, 149 Nm/s→ 33, 149

745.7= 44.5 hp

(10)

Repeating the above calculation with d = 0.001 mm,

FAvg =1.47

d=

1.47

0.001× 10−3= 1, 470, 000 N (11)

Thus the acceleration is a = −9, 800, 000 ms−2 with a duration of t = 452.04 ns. Thus,

1.47

452.04× 10−9= 3, 251, 918.74 Nm/s→ 3, 251, 918.74

745.7= 4, 360.9 hp (12)

From the above analysis it can be seen that dropping an object (with a mass of 150 grammes) from a heightof 1 m (∼3 ft) will cause it to experience a shock of ∼99,900 g whilst generating ∼45 hp of power, if thedistance travelled during impact is '0.01 mm.

However, the above analysis does not take the ‘crumpling’ effect of the object into consideration. Uponimpact, this could further increase the distance over which the object decelerates. If the metal object inquestion were to be dented by, a modest, 1 mm it would only experience a deceleration of 998.98 g whilstabsorbing ∼0.44 hp.

To put this into perspective, it is safe to assume that a 3.5” computer hard disk drive (HDD), with anapproximate weight of 0.7 kg, falling through a height of 1 m onto a solid surface would experience adeceleration of 1000 g. The crumpling distance would be much smaller, in this case, for side-on collisions.