Apple granted patent for flipping an iPhone in free fall, so it doesn't land flat on its face

Apple granted patent for flipping an iPhone in free fall, so it doesn't land flat on its face
Those contraptions for preventing smartphone drop damages are starting to shape up much better now than their initial variations which include airbag deployments to cushion the blow, for instance. Straight from the US patent office comes Apple's latest grant that deals with automatically protecting your iPhone during a free fall.

Thankfully, this one doesn't include any airbags stuffed inside the handset, but rather engages the sensors that are already in the handset anyway - accelerometer, magnetometer and the like - for determining quickly the side and point of impact. Afterwards, a dedicated processor calculates the needed adjustment for a fraction of a second, and engages a motor with eccentric mass to adjust the phone's position mid-air, and land it on its least breakable side, instead of face-down on the concrete, for instance. 

In fact, the vibration motor that is already in the handset, can be used for the purpose, just with stronger vibe impulses, compared with the ones that are used to alert you for messages or incoming calls when the phone is on mute.

source: USPTO via AppleInsider



1. mercorp

Posts: 1045; Member since: Jan 28, 2012

How about making sure that it dosen't bend first?

15. power_x

Posts: 264; Member since: Aug 28, 2013

They have a solution It's called don't put your $700 device in your back pocket sitting on it

21. jaytai0106

Posts: 1888; Member since: Mar 30, 2011

There are report that people bend their phones when they are in the front pocket though. However, this might be one thing that is truly innovating coming from Apple in a long time.

30. wilsong17 unregistered

how come i put my note 3 on my back pocket and it doesnt bend

36. dirtydirty00

Posts: 428; Member since: Jan 21, 2011

i thought everyone was looking foward to a flexible phone???

43. InspectorGadget80 unregistered

How bout using IP RATINGS instead of filing for patents and use it to sue others later one.

2. tech2

Posts: 3487; Member since: Oct 26, 2012

Seems like Apple took last year's PA's joke literally and really started working on a feature that would prevent damage to phone from falling:

5. Droid_X_Doug

Posts: 5993; Member since: Dec 22, 2010

This is fair game for patentable innovation - technical solution to a problem. Props to Apple.

47. maherk

Posts: 6750; Member since: Feb 10, 2012

Nah, they ll just patent it and wait for someone else to make it happen. Then they come with their lawyers to make sure they take your invention and make you pay millions on top of that.

3. rantao333

Posts: 345; Member since: May 21, 2013

yes, i definitely wish to have this feature!

4. uchihakurtz

Posts: 427; Member since: Nov 12, 2012

The ladies are gonna like jumping around if this was in the new iPhone

6. vuyonc

Posts: 1089; Member since: Feb 24, 2014

Airdrop version 2 :P

7. darkkjedii

Posts: 30902; Member since: Feb 05, 2011

Not bad Apple, not bad at all.

8. NexusPhan

Posts: 632; Member since: Jul 11, 2013

I doubt this would work practically. The inertia of a heavy phone falling is too much. I took a look at the phone falling face first. F=mg = 172 grams (iPhone 6+) * 9.8m/s^2 = 1.686 kg m/s^2 To counter-act this inertia, we have to balance this with the centripetal force equation, F=mrw^2 (w= lowercase Omega, m = the "eccentric mass" ) = 1.686 kg m/s^2 This thickness of a 6+ is 7.1 mm, so the maximum radius this could ever be is about 3.5mm (.0035m). Let's call the angular velocity 1,000RPM or about 100 radians/second (read: FAST) so F = m * .0035 * 100^2 = 1.686. Solving for m, m = .048 kg = 48 grams!!! 48 grams is so substantial compared to the overall weight of the iPhone, we would have to iterate to get the correct, heavier weight! Unless Apple has something entirely new (doubt that at this point), this just does not seem feasible to me.

10. rowdysheeter

Posts: 74; Member since: Sep 10, 2014

thanks for the explanation.

14. Leo_MC

Posts: 6935; Member since: Dec 02, 2011

Speed. If you put the speed into the equation, the results will be different.

18. NexusPhan

Posts: 632; Member since: Jul 11, 2013

Speed of what? The electrical motor? It's in there and estimated at 1,000 RPM which I don't think is possible to reach in the less than 1 second it would take for the phone to drop 2 meters.

20. Leo_MC

Posts: 6935; Member since: Dec 02, 2011

People didn't think it was possible to walk on the moon.

23. NexusPhan

Posts: 632; Member since: Jul 11, 2013

Ignoratio elenchi. Google it, if you need to. In English, we'd call it a Red Herring. Maybe Google that too.

40. Leo_MC

Posts: 6935; Member since: Dec 02, 2011

The calculus ignores some variables; I don't know physics well enough to explain it in English.

41. NexusPhan

Posts: 632; Member since: Jul 11, 2013

I looked at worst case, which is generally how you should do it. Look below (post 33) for a more realistic approach that yields a similarly unpractical mass requirement.

42. Leo_MC

Posts: 6935; Member since: Dec 02, 2011

For rotating corps F (vect) = mvr (vect)/r. So, the force is proportional with the speed (the higher the speed, the bigger the force).

24. duartix

Posts: 311; Member since: Apr 01, 2014

Thank you for the report! :) Why do you limit the radius to 3.5mm? You can have a much bigger centripetal forced if aligned with the plane defined by the screen. You just can't make the phone turn in a 3D space though. But maybe that's not needed for the protection effect and you can just relax this major constraint. What are your thoughts on this?

27. NexusPhan

Posts: 632; Member since: Jul 11, 2013

I assumed this mass would be required to rotate in any direction, depending on the phones exact orientation when falling. In order for the eccentric mass to rotate, the diameter must fit within the phone's thickness. A 7.1mm phone thickness meant a max 3.505mm radius (I rounded for the calculations). I think this will be really, really tough to make practical. They need to focus on shatter proofing the screens. Sony had these (Xperia Z1) but people hated the coating that was required to make them shatter proof, so Sony stopped using it. Improve the coating and bring it back!

54. duartix

Posts: 311; Member since: Apr 01, 2014

I'm thinking a bigger challenge will be to determine how far from the floor the phone is. Unless of course they can easily correct the trajectory for overshoot correction inertia once the phone has achived an horizontal fall on it's back, which I assume is the optimum. BTW, 2m isn't enough distance. I believe the median drop height is probably more close to 1m, which will demand for twice the reaction speed. :(

25. imnotascammer

Posts: 160; Member since: Aug 22, 2012

Whew. Nerd alert! Lol. Just kidding man.

28. NexusPhan

Posts: 632; Member since: Jul 11, 2013

I'm an aerospace engineer designing plane, rocket and satellite components. (Been to Space X which was freaking awesome BTW) These calculations are what I do every day. I am a super nerd and proud of it!

32. TournaFlyer

Posts: 27; Member since: Sep 20, 2014

I understood every word of that...about 3rd grade level where I come

33. vuyonc

Posts: 1089; Member since: Feb 24, 2014

It would be nice if the comments system allowed free body diagrams. Anyway, equating gravitational force to a counter-acting centripetal force doesn't seem to make sense. Wouldn't rotational motion over time be the controlled factor to protect the face? Mass moment of Inertia of Iphone 6+ along its height for example (I) = 1/12 * mass * (width^2+thickness^2)=1/12 * (.00778^2+.00071^2)= .00605 kg.m^2 Assuming the fall takes 1 second (t), initial angular velocity being zero, the phone spinning perfectly around (angle theta = pi radians), Theta =.5 * alpha * t^2 From there alpha = 2 * pi = 6.28 rad/s^2 The counter-acting torque (T) = inertia * alpha = .038 Nm Taking your angular velocity of 1000rpm (105 rad/s) The minimum angular acceleration of the contraption (alpha2) = 105 rad/s^2 The minimum inertia of the contraption = torque / alpha2=3.62 X 10^-4 kg.m^2 With your radius of 3.5mm, the mass = 10 grams But then you would need to account for gyroscopic torque if the phone's already spinning. Meh, negligible in small masses. Too lazy to continue.

34. NexusPhan

Posts: 632; Member since: Jul 11, 2013

I complete agree. I tried to look at a worst case scenario where the phone has to completely overcome the entirety of the gravitational force. In real life, this would never be the case. I took the lazy man's approach. Thanks for your additions. But, I think even at 10 grams it would take up too much space and weight in the phone to be feasible.

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