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You may have seen the story of the iPhone which was dropped from perhaps 13,500 feet by a skydiver - it survived.

This made me wonder how to work out the terminal velocity for something like that. Obviously calculating terminal velocity for a sphere can be relatively straightforward, but with a flattened oblong, what factors come into play?

End on will be fast, flat will be slow, but is there a stable configuration?

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I can tell you that there is no stable configuration. It might be helpful to assume that it is stable though. –  JoeHobbit Jul 19 '11 at 14:15
    
I sent @Rory that link in The DMZ (security.se's chat room). I'm also interested in this. If we assume that it is stable, what factors would come into play? –  Thomas W. Jul 19 '11 at 14:18
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I wonder if there's an app for that? –  Iszi Jul 19 '11 at 14:18
    
As for the survival, I assume it did help that "The iPhone had protective gear of its own -- an Incipio-brand phone case that was broken after the fall but still was on the phone." –  nealmcb Jul 19 '11 at 14:43
    
""Obviously calculating terminal velocity for a sphere can be relatively straightforward,"" Not at all. Not at such Reynolds numbers. –  Georg Jul 19 '11 at 14:45

2 Answers 2

up vote 3 down vote accepted

From the equilibrium between drag and weight: $$ \frac{1}{2} C_x \rho v^2 S = m g $$ we can write the terminal velocity as $$ v = \sqrt{\frac{2 m g}{C_x \rho S}} $$ where $m$ is the mass of the phone, $C_x$ its drag coefficient, $S$ its section, $g$ the acceleration of gravity, and $\rho$ the density of air.

Now, this is not really a good answer, as the big question is how to estimate $S$ (depends on the phone orientation) and $C_x$. But at least you can compute an order of magnitude, as in most cases $C_x$ of of order 1.

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For something nearly square $C_x$ is going to be over 0.5 and nearly 1.0. Also the phone twists as it falls, and so the swept area changes all the time. –  ja72 Jul 21 '11 at 0:54
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@ja72 Ah, You know the phone will twist? –  Georg Jul 21 '11 at 12:49
    
@Georg - I meant twirl ! –  ja72 Jul 21 '11 at 18:08
    
All the same! I guseed You meant twirl. –  Georg Jul 21 '11 at 18:18
    
I am like some people, a bit lazy. What's the answer? the iPhone 4 weighs 0.137 kg –  Alvar Jul 21 '11 at 18:49

Suppose that there was no turbulence in the air.
It will fall down as a leave, horizontal, slightly oscillating around he vertical axis in relation to the Earth surface, about 6 degrees, the same as the long ice plates do within contrails and cirrus clouds.

In the formula of Bonet's answer the section S is maximum (good to minimize the speed). The drag coeff ? I dont know. Someone can edit this answer to make it more complete.

But the physics do not end here, I think. The phone will slide in large horizontal moves, as a sail or wing, much like a skydiver do. It is much more than a simple vertical ballistic motion.
Imo, thats the why.

EDIT add
assuming only an horizontal descent (the mass distribution is centered) :
exploring the numbers: the best result, for Cx = 1.3, is v= 15.6 m/s (in my oppinion it is too much)
Cx 0.5; 1; 1.3 ; g 9.8 ms^(-2); Ipod4 137g, dept 9.3mm surface 0.1152*0.0586 m2; Density of air 1.25 Kg/m^(-3)
Cx=0.5 v=25m/s; Cx=1 v=17m/s; Cx=1.3 v=15.6 m/s

We have to explore other aerodynamic effects.
altitude 4100 m , retrieved 800m away from where he landed. The angle of the descent of the Iphone4 probably is greater than 11º (arctan(800/4000)=arctan(1/5)). From this we can conclude that it was not a balistic descent and aerodynamic effects indeed are present.
I do not know the mass distribution in the Iphone4. If it can be made similar to a marble seed, marble seed
with much of the mass near the one top then the cellphone will rotate and significantly slow down.
reference EXPLORING THE BIOFLUIDDYNAMICS OF SWIMMING AND FLIGHT by David Lentink pages 111-119 chapter 5.1 A LEADING EDGE VORTEX PROLONGS DESCENT OF MAPLE SEEDS
The package ELMER, among others, can be used to make the numerical aerodynamic simulation with all this variables including, if needed, the building and the wind (any upward air flow in the trajectory of the mobile ?).
EDIT add end

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According to the Wikipedia article on drag coefficient, if we assume the phone stays horizontal, $C_x$ should be something like 1.28. I don't know how much we can trust those numbers though. –  Edgar Bonet Jul 20 '11 at 19:39
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The example of cirrus ice platelets is nonsense! This happens at very small Reynolds numbers! The cell phone may fall "flat" (maybe oscillating) or may tumble, I dont know. But always at Reynolds Number much too big to allow "calculation" of velocity. Best one can achieve is some estimation. –  Georg Jul 21 '11 at 12:47
    
@Georg quoting from Atmospheric Optics site "Plate crystals drift down like leaves. Their large faces are almost horizontal." and [also here] (atoptics.co.uk/halo/orplate.htm) (about the 6 degrees I cant recall from where I got it). Answering this question I try first to find ways to minimize the vertical speed. I think that a numerical simulation is the best way to make an approach to find the range of a viable terminal speed. –  Helder Velez Jul 21 '11 at 13:56
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@Helder Can You read? Nodody doubts that those ice leaves behave that way. (This can be reproduced with small pieces of paper!) But this is in the domain of very small Reynolds numbers! Do You know what that means? –  Georg Jul 21 '11 at 15:05
    
@Georg I will not answer to your 1st and last ?s . Why did you wrote: The example of cirrus ice platelets is nonsense! IMO they have the same stabilization process: "small deviations produce correcting forces that restore the orientation." (previous given link). You have 4 comments in the Question/Answers. Can we have a answer of your own explaining why the phone survived? and an estimate of the terminal speed, the Reynolds Number, and considerations about the use/(not use) of DNS, LES,RANS,DES methods ... if you please. –  Helder Velez Jul 21 '11 at 18:11

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