# Surviving a fall into a balloon-filled pit [closed]

My brothers and i have been debating this question for years. If an average person fell at terminal velocity into the center of a pit that is 1 mile deep x 1 mile in diameter, and the pit is filled with typical latex-based air-filled round party balloons (not helium-filled), would they survive the impact? Each balloon is independent of the other balloons.

Thanks

Edit: this only concerns the impact, not the subsequent events that occur after the person has stopped.

Edit: there are several ways to die from the impact - too abrupt of a stop (do the balloons give enough?), not enough of a stop (would you fall through the balloons? Do they pack densely enough?), static electricity (?), heat (?), etc.

• FWIW, terminal velocity is around 122 mph (53 m/s). If you assume the "average person" weighs about 180 pounds (82 kg) the person has a kinetic energy of 1/2 m v squared -- 0.5 28 2809 = 39,326 joules (if I didn't screw up the math). – Hot Licks Dec 29 '18 at 1:02
• Doesn't it depend on how filled the balloons are and how densely packed they are? – Aaron Stevens Dec 29 '18 at 1:06
• Aaron, yes. Consider typical party balloons – Paul Stonaha Dec 29 '18 at 1:51
• All Nicholas Alkemade needed was some snow and a few pine trees... en.wikipedia.org/wiki/Nicholas_Alkemade – DJohnM Dec 29 '18 at 3:52
• Paul, there is an edit history for those who are interested. It would be better to make one cohesive question instead of explicitly including an edit history as part of the question. – Aaron Stevens Dec 29 '18 at 5:21

Eventually, the person falling into the balloons would build up a "pyramid" of tightly-packed balloons under him. The balloons in this pyramid would compress until they popped. When they pop, the energy that was put into them to compress them would be mostly dissipated without any being redirected to our unlucky skydiver.

The trick is figuring out how much force a given balloon will absorb before it bursts. Apparently the typical party balloon pressure is only about 1 psi greater than atmospheric, and one could make a SWAG that the pressure would get up to 3 psi (0.21 kg/cm2) when it bursts. If the balloon has a radius of 15 cm then it has a surface area of about 2800 square cm. Guestimating that about a third of the surface would be in contact with the diver or adjacent balloons when it reaches max pressure, the force produced by the almost-burst balloon would be 0.21 * 2800 / 3 = 196 kg. But only half this force would be directed upward, so figure 98 kg.

This number is clearly wrong -- it would mean you could stand on the balloon without bursting it if you weren't too heavy. But most of us have run this "experiment" to one degree or another in our younger years, and, for a reasonably tough balloon, a number of 5-10 kg is not unbelievable.

The other issue is how much motion occurs as the balloon is compressed and what the force/distance curve looks like. Again, difficult to estimate, fairly easy to measure. But let's assume that the net effect is that, during the last 5 cm of motion, the force remains constant at that 5-10kg number. So in 5 cm distance the balloon absorbs 0.25-0.50 kg meters of energy. Or (1kg-m = 9.8 joules) about 2-50 joules. Let's say 25.

From earlier:

FWIW, terminal velocity is around 122 mph (53 m/s). If you assume the "average person" weighs about 180 pounds (82 kg) the person has a kinetic energy of 1/2 m v squared -- 0.5 28 2809 = 39,326 joules (if I didn't screw up the math).

Let's So it would take bursting 39,326 / 25 = 1572 balloons to absorb all the energy in our unlucky victim. The balloons are 30 cm in diameter, so a stack of them one mile high would be 1609 / 0.3 = 5363 balloons -- about 3.5 times the minimum needed.

So, if things lined up correctly, and all the thumb sucking above is not too terribly wrong (and no major math errors) then it's plausible that your mile-deep lake would be sufficient to stop the fellow from bashing his head on the bottom of the lake.

Of course, the balloons will not stack neatly on top of each other but will arrange themselves like grains of sand in a pile. Some effects of this would be bad, others good.

And this has gone on far longer than I wanted.

• Did you take into account that inside the pit the skydiver not only loses kinetic energy in collisions with balloons but also gains kinetic energy due to gravity? – akhmeteli Dec 30 '18 at 4:51
• @akhmeteli - I'll let you do the computations on that part. – Hot Licks Dec 30 '18 at 13:03
• OK, here you are. The kinetic energy that the skydiver gains due to gravity while moving to the bottom of the pit is $W=m g h$, where $m=82 kg$, $g\approx 10 m/s^2$, $h\approx 1600 m$, so $W\approx 1300000 J$. So maybe the word "plausible" in your answer should be replaced by "implausible". – akhmeteli Dec 30 '18 at 16:59
• And, by the way, the kinetic energy of a 82 kg person moving at the speed of 53 m/s is $82\cdot 53^2\cdot 0.5\approx 115000 J$. – akhmeteli Dec 30 '18 at 17:12
• Yep, I apparently transposed 82 into 28. – Hot Licks Dec 30 '18 at 19:35

The situation you're describing is a high-energy impact with a granular material.

While certainly some energy is going to be expended by popping balloons, I think friction amongst balloons and between yourself and the balloons is going to be the dominant effect.

Hot Licks's answer is basically describing a force chain. Modeling a force chain in a mile cube of latex balloons is probably going to take a largish computer⸮

I find akhmeteli's answer unconvincing. In particular, I don't think it's reasonable to model a mass of balloons as having the same viscosity as unconstrained air.
(As an aside, if you are trying to measure the "viscosity" of a mass of balloons, make sure to model them as a shear thinning fluid.)

Granular material is complicated. I see two paths toward a satisfying answer, and I recommend them both.

• Start whatever further schooling you need to become a research physicist. If you tell them you want to study granular materials I think you'll find funding; there are lots of industrial and civil-engineering applications.
• Do an empirical test. The quantity of balloons you'll need for a small demo is staggering, but not unprecedented. Hitting the balloon-pit with a sandbag thrown from a helicopter is probably harder than it sounds; if you think you have enough balloons you could just jump and steer a bit on the way down.
• I find much of what you're saying reasonable. I think one needs to come up with a model of the effective viscosity of a pit of balloons to properly solve this. Actually testing this experimentally, even a model, is not practical; because the pit is a mile wide, edge effects will be negligible (open boundary conditions). – Paul Stonaha Dec 30 '18 at 4:03

In order to answer your question, let us compare an air balloon and a volume of air in the atmosphere having the same shape as the balloon. According to https://www.youtube.com/watch?v=fwh-i0WB_bQ , the pressure does not differ more than by 10% in these cases (at a higher excess pressure the typical balloon pops). Next, the mass of the balloon envelope is less then the mass of the air it contains (as the same balloon floats if filled with helium). Therefore, the average density in the pit filled with balloons should not exceed the density of atmospheric air more than by a factor of 2. Then, using the drag equation for a rough estimate, we assess that the drag in the pit should not exceed the drag in the atmosphere more than by a factor of 2 at the same velocity. Thus, the terminal velocity in the pit should not be less than the terminal velocity in the atmosphere divided by $$\sqrt{2}$$. Thus, the chance of survival for a free fall in the pit should be close to zero. Let me emphasize that gravity is still very much the same in the pit. If there were no gravity in the pit, the skydiver would have slowed down to a safe velocity, but the same would be true for a pit filled with air if there were no gravity in the pit.

• Thank you for the answer. I don't think your answer properly considers the dynamics of the balloons, however. I believe there should be an elastic restoring force from the balloons' compression. Or at least, we need to consider interballoon dynamics. – Paul Stonaha Dec 29 '18 at 5:51
• @PaulStonaha : elastic forces cannot provide extra pressure in excess of 10% (balloons pop at such excess pressure), so elastic properties of balloons should not have significant effect. – akhmeteli Dec 29 '18 at 6:00
• Akhmet, if you fall onto an inflated balloon and pop it, you'll feel a resistance before it pops. This is due to an increase in the balloon's air pressure, along with the increase in elastic potential energy of the latex. If the balloon were inflated to its "popping pressure" (+10%, as you say) the slightest compression would pop it. That is not how party balloons are inflated, in practice. – Paul Stonaha Dec 29 '18 at 16:01
• @PaulStonaha : Let us consider an example where a man steps on a balloon with his two feet. Let us assume that the surface of the soles is 150 cm^2 (based on en.wikipedia.org/wiki/Ground_pressure and man's weight of about 80 kg). Atmospheric pressure times 10% times 150 cm^2 is about 15 kg-force. So yes, a balloon can provide a lot of resistance. However, this does not seem relevant, as we get such resistance if we step on a balloon lying on the ground. In your case, however, balloons are not on a solid surface, so they will move away, and resistance will be much smaller. – akhmeteli Dec 29 '18 at 16:44
• Akhmet, I'm still not convinced. How quickly will the balloons move away? – Paul Stonaha Dec 29 '18 at 22:20