# How big of a hampster ball would I need to survive a fall from an airplane?

There is a pit threat right now discussing the merits of putting kids in hampster balls to protect them from life.

My question: Suppose I wanted to sky dive w/o a parachute. How big would by Hampster ball have to be for me not to break a leg on impact. suppose the ball were something like the Zorb Ball

How big would one of those have to be for me to fall out of an airplane, bounce a few time on impact, and live through the whole thing

Thanks

I have no real clue on exactly how big the ball would need to be, but I do know the terminal velocity of humans is around 120MPH… so I would think that if you were in a big ball, it would be quite a bit slower. You might not ever exceed 80 MPH with that much ball around you creating wind resistance. And if you were only going 80, then you would be going about the same speed as a car crashing at high speed.

So, I would think if you were secured very well so your neck and limbs don’t wip around, and the ball did not burst upon impact, maybe a 25 foot ball would be ok. There are too many variables here though - ball stength, elasticity, internal packing, restraints, etc. But if we can land on mars with airbags, and humans can (sometimes) survive an 80MPH car crash with a seat belt and an air bag, I would think it could be done.

It would be quite entertaining to see big balls dropping out of airplanes and bouncing a long the earth. That would be a cool extreme sport.

It would take some very big balls, indeed.

You can be absolutely certain that if it were physically possible to launch those things from a plane that someone, somewhere has turned it into an extreme sport. I’ve seen it in movies before, something with Jackie Chan, one of his low budget, less successful films.

I think the drawback here would be finding a suitable plane to launch these monster balls from. Most skydiving planes are fairly small, and finding a big plane to do this from may make it to costly to do.

Maybe a helicopter could lift the big balls up, 2, 3, or 4 at a time and drop them over a safe area? But helis are expensive to run as well and have a lower flight ceiling.

Maybe it could be quickly inflated during the first few moments of descent (admittedly it would build up a bit more speed that way). I think we could safely say that something the size and more or less the weight of a hot air balloon envelope would work really well (as long as you could keep it fully inflated and stay inside, preferrably suspended nicely near the top (balance is a problem though).

A parachute presents sufficient wind resistance to allow a safe descent, but that again is a lightweight structure; a Zorb is quite a bit heavier, so it would have to be bigger, whihc in turn would make it heavier etc; but there would be a break-even point, I think.

There was a design for an emergency re-entry vehicle that was just like a big cone, with the occupant sitting in a very well-padded chair, jammed arse-rirst into the apex; I don’t think it was ever built or tested though.

What kills you when you hit the ground sans chute is not the impact but the body’s inability to absorb all the G’s caused by said impact. While the hamster ball is going to absorb some of the G’s of the whole contraption hitting the ground, will it absorb enough G’s to keep the occupant from getting hurt when he absorbs G’s from hitting the inside of the hamster ball?

I would bet that in a vacuum (so that terminal velocity is actually reached, the ball would have to be extremely (ind impractically) huge.

There’s no p in the word hamster.

Forget the ball, just make sure there’s some power lines in between on the way down.

Would it help if we made it a giant Buckyball with shock absorbers at each intersection?

'Cause, you know, that would be neat.

Just thought I would bring up one of my failed predictions, which was that once airbags in cars were commonplace, a new thrill-seeking sport would develop, namely one in which two participants would crash their cars into each other head-on at 50 mph. WOW WHAT A RUSH! I guess it’s too expensive though, as would (I think) the giant-hamster-ball skydiving imagined in the OP.

I don’t know but I heard on MSNBC today that a woman parachutist survived an 11000 ft fall in which both parachutes failed. They apparently opened enough to slow her down so that when the chutes snagged a power line her fall was further slowed enough to merely, if that’s the word, give her a cracked pelvis.

I’ll look for a cite.

Damn! I missed** II Gyan II’s** post and thereby wasted memory.

I agree with Mangetout, there should be a break even point. It would probably have to be pretty large though, and the person would have to be suspended in the center with some sort of elastic membrane. For example, take the zorb and scale it by a factor of 10. Now make a cris-crossing system of elastic/rubbery cords that keep the person centered in the sphere (perhaps we should connect the cords to a human-shaped harness instead of to the human themself). With any luck, the cords will be rigid enough to absorb enough G’s to keep the human alive.

IIRC, you can use Stokes’ Law to figure out the viscous force on the sphere. If i get some time later tonight or this week i’ll see if i can’t hash out a solution to this problem.

Stunt folks that jump off of buildings for a living now have materials/devices that will let you survive a fall from any distance–at terminal velocity. It would seem that one could wrap such a thing up into a ball…

(Couldn’t find a cite or site for the “fall from infinity” stuff, but I know I saw it on TV, so it must be true.)

OK, here’s a quick back-of-the-envelope calculation for you. Stokes’ Law states that the drag force on a sphere of radius a traveling through a medium of viscosity η at velocity v is:

``````
**Fd=6πηav**

``````

Now, for the assumptions:

1. We’re going to initially ignore the buoyant force of the air on the sphere, which we will find to be an unsatisfactory approximation.
2. we’re going to ignore the fact that the viscosity of air changes with height, and approximate it using the value at ~15 degrees C
3. We’ll assume the restraints are solid enough to keep a human alive through the course of an 80mph collision with a stationary object (the earth). thus, we let
``````
**Fg=mg=Fd=6πηav**

``````

and solve for a:

``````
**a=mg/(6πηv)**

``````

We’ll assume the sphere is light, and weighs about the same as the human, so the mass is twice the mass of an average 150lb human. Thus the values we plug in are:
m=136 kg
g=9.8 m/s^2
η=1.810^-5 Ns/m^2 (viscosity at 15C)
v=80mph=36 m/s

This gives us a radius of a=1.09*10^5 m = 109 km. This seems unreasonably large. The first thing we note is that at this size, the buoyant force is almost certainly not negligible. In fact, the buoyant force Fb would be

``````
**Fb=dg4πa^3/3**

``````

using the value of the density of air (d) at sea level (1.29 kg/m^3), this results in a buoyant force of 6.9*10^16 N, which is huge compared to the drag force (1331 N).

So, to revise our model, we include the buoyant force in the equation:

``````
**Fg=Fd+Fb

mg=6πηav+dg4a^3/3**

``````

plugging in the values given above and solving for a in Maple yields the value a=2.93 m, a much more reasonable number, though it seems a bit small. It’s worth noting that since d will decrease with increasing height, this value of a is bound to be a low estimate. However, it is conceivable to make a sphere 10x larger than this (~30 m), though it would be difficult to hoist into the air. Furthermore, this is assuming the harness structure is designed properly, so that it reduces the impulse (G’s) applied to the human by enough to keep them alive.

I’d like it if someone wanted to check my calculations – while i trust Maple, i don’t trust my ability to enter everything into it correctly.

OK, 120 MPH is about 54m/s lets say that the drag of the ball slows the terminal velocity to 50 m/s. Further, assume the max acceleration you want is -5g (one can obviously survive higher, but I’m trying to find a balance between “really safe” and “emergency use only”)

going from 50 m/s to 0 m/s at 50 m/s^2 takes 1 second (I think, its late)
distance = 1/2 at^2 = 25m, hmm , thats pretty big
at 10g thats .5 second = 2.5m – more reasonable size

(note, this is the size in between the 2 spheres of the zorb ball, the part that compressess)

Brian
*again, it is late so sorry if I misremebered equations or did the math wrong

Imagine if you will…probably quite doable. Make the Bubblechute[sup]TM[/sup] an inflatable device that is folded into a backpack similar to a regular parachute. Pull the cord and a small tank of compressed air inflates the ball.

Considering that a quick deflate valve of some kind could be pretty easily rigged, it might make a better parachute arrangement than present day ones. As a nifty bonus you might even be able to make them from bullet resistant fabric. By design they would probably also fall faster than normal parachutes reducing exposure to ground fire. So why isn’t the millitary using them already…

You can do a quick approximation easily using two factors:

1. your own assessment of the maximum tolerable deceleration (let’s say 8G or 78 m/sec²)
2. the terminal velocity of a large sphere (let’s say 80 mph or 36 m/sec)

If we assume that the Zorb ball provides a constant deceleration, then it only needs to have a radius sufficient to decelerate a human from the chosen velocity at the chosen rate.

For 8G deceleration from 80 mph:
T = V/a = (36 m/s) / (78.4 m/s²) = .46 sec
D = at²/2 = (78.4 m/s²) * (.46 s)² = 16.6 m

Add 2m for the inner sphere for the Zorb ball. In principle, that’s all you need, regardless of how you do the caluculation - though of course the terminal velocity above is a crude guess. My back of the envelope calculations actually give a lower number.

Designing a Zorb ball that will assure a constant deceleration is the challenge, but not an insurmountable one. You may need a few extra meters for the spring network and a mechanism to keep your landing chair in a constant orientation (you wouldn’t want to be spun to death in a rapidly rolling ball, though the larger the sphere, the less rapidly it would roll) The constant orientation of the inner sphere (or chair) would also allow you to assure that you would land on your back, which is probably the safest position for uniformly distributing the force over your entire body, and preventing injuries.

The overall experience would comparable to a jet pilot doing an 8G inside turn, but it would have a much more rapid onset, and the force would be perpendicular to your axis, while a pilot’s head is always somewhat above his head. If you wear a G-suit and do the straining maneuver, you stand a good chance of being able to emember the experience rather than blacking out, and emerging with no real injury.

D’OH. I got called away from the computer. Someone beat me to it!