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IPhone 4 Survives 1,000 Foot Fall From Plane
tekgoblin writes "From the article: 'US Air Force Combat controller Ron Walker had lost his iPhone 4 from his aircraft during flight. He works as a Jump Master, which is where he would ensure the airplane was in the correct position when he sends parachute jumpers out. The plane was moving at 150 mph and while looking out the door of the plane to find necessary ground landmarks his pocket opened and his iPhone flew out. When he noticed his phone fell, he thought all was lost. Upon landing and sharing the story with friends he installed the Find My iPhone app on one of their phones and went looking for his phone. He expected it to be battered from the fall but found the phone to be 100% un-damaged from the fall. The phone was protected by a Griffin Motif TPU iPhone case but it isn't clear whether the case protected the phone from the fall or the fact that it was cushioned by the brush that it hit.'"
That WAS the article, minus the last sentence.
So we know now that the iPhone 4 can survive a 1000ft fall as long as it doesnâ(TM)t hit concrete, I wonder if Apple will talk about this at one of their next iPhone announcements.
There, now you've read the entire article.
I always knew that the Chinese manufacture the best equipment in the world.
Small objects have proportionately more drag for their weight so their terminal velocity may not be that fast, reached earlier (so overheight doesn't matter) and damage less.
Another case of why there are no flying pigs -- weight increases as the cube of length, while drag increases as the square. So lots of flying bugs.
It would read 1G after it had hit the ground (stationary lying on the ground), but during the impact, it would be far more than that. It would read 0G during freefall (as this is the definition of freefall). The deceleration as it hits the ground would be very high, (possibly even as much as 100G+) because the time to decelerate is so small, and the distance over which it decelerates is tiny.
You would need to perform controlled tests under identical conditions to decide which is the more durable in reality. Please post the results on YouTube.
From scarped cliff or quarried stone she cries "A thousand types are gone, I care for nothing, no not one."
The Gs will change through the flight
When it initially starts falling air resistance will be negligable and it will experiance approximately 0G
As it approaches terminal velocity and stops accelerating the G-force experianced will increase tending back towards 1G
When it hits the ground things get complex
After it has settled on the ground it will experiance 1G
During hitting the ground is where things get really complex. A simple model assuming that the objects are rigid and that "contact forces" appear instantlygives a result of infinite acceleration and therefore infinite Gs. In reality different parts of the object will experiance different Gs as the object deforms on impact.
note: i'm known as plugwash most places but i screwd up registering that here somehow in the past and now can't register
Well, if there's a landing strip where the bush has been cleared, aim for that.
If it's wide open space instead, give thanks for your good fortune and aim for the open space where the bush used to be.
Terminal velocity is calculated as sqrt((2*w)/(rho*A*Cd)). .07 ft**2 .75
w = weight
rho = about 1.22 kgm**3
A = Things fall in the orientation that causes the most air resistance (believe it or not) so that's the face area, about
Cd = Coefficient of drag for a rectangle is about
So terminal velocity is about 50 MPH.