# MythBusters: How small could a lead balloon be?

On a previous episode of The MythBusters, Adam and Jamie made a lead balloon float. I was impressed. Anyway, I decided to give a more detailed explanation on how this happens. Using the thickness of foil they had, what is the smallest balloon that would float? If the one they created were filled all the way, how much could it lift?

First, how does stuff float at all? There are many levels that this question could be answered. I could start with the nature of pressure, but maybe I will save that for another day. So, let me start with pressure. The reason a balloon floats is because the air pressure (from the air outside the balloon) is greater on the bottom of the balloon than on the top. This pressure differential creates a force pushing up that can cause the balloon to float.

**Why is the pressure greater on the bottom?**
Think of air as a whole bunch of small particles (which it basically is). These particles have two interactions. They are interacting with other gas particles and they are being pulled down by the Earth’s gravity. All the particles would like to fall down to the surface of the Earth, but the more particles that are near the surface, the more collisions they will have that will push them back up. Instead of me explaining this anymore, the best thing for you to do is look at a great simulator (that I did not make)

![Page 0 Blog Entry 14 1](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-1.jpg)

When you run the simulator (a java applet) you will need to add some gas in the chamber by moving the handle on the pump. When you do you will see that there many more gas particles at the bottom of the container than at the top. If you look at the balloon inside the chamber, there will be more particles hitting the balloon from the bottom than from the top. Since there are more collisions on the bottom, this creates a total force from the collisions pushing the balloon up. How would one calculate how much this force is? Well, the simplest and sneaky way is the following: Suppose I did not have a balloon there at all, but there was just more air. What would that air do? It would just float there. Here is a force diagram for some of that air:

![Page 0 Blog Entry 14 2](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-2.jpg)

So, the forces have to be the same (gravity and the force from the collisions – also called the buoyancy force). If these forces were not the same, this section of air would accelerate up or down. Yes, the density of this air is not constant, but that doesn’t matter. Thus (I like saying thus) the buoyancy force must be equal to the weight of this air.
Now put a balloon (or any object – like a block of pudding) in that same space. The gas around it will still have the same collisions resulting the same buoyancy force. This is where Archimedes principle comes from that says “The buoyancy force is equal to the weight of the fluid (or air displaced)”

This principle can be written as the following formula:

![Page 0 Blog Entry 14 3](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-3.jpg)

Where ? is the density of the stuff the object is in (in this case it would be air). g is the local gravitational constant – that turns mass into weight. V is the volume of the object.

**Here is the data from the MythBuster’s balloon.**
I wrote down the dimensions of the huge (ginormous) balloon from the last episode. Here is what I have to start with:

• mass of lead used = 11 kg
• surface area of lead used = 640 ft2 = 59.5 m2 (from google calculator – just type “640 ft^2 in m^2”)
• Also, they say it will have 30 kg of lift (which isn’t technically a proper thing to say, but if I take this to mean 30 kg *9.8 N/kg = 294 Newtons – then ok)
• They also claim the balloon will be a 10 ft by 10 ft by 10 ft cube. If that was the case, it would have a surface area of 10*10*6 = 600 ft2. I guess the extra 40 square feet is from overlapping material.

**How thick is the foil?**
The density of lead is 11,340 kg/m3. Here they have a rectangular solid that looks like this:

![Page 0 Blog Entry 14 4](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-4.jpg)

Such that it has a volume of:

![Page 0 Blog Entry 14 5](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-5.jpg)

I know the area already. The volume can be found from the mass (and the fact that it is lead). Density is defined as the mass/volume so:

![Page 0 Blog Entry 14 6](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-6.jpg) and ![Page 0 Blog Entry 14 7](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-7.jpg)

This would mean that the thickness would be:

![Page 0 Blog Entry 14 8](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-8.jpg)

That’s pretty thin. This is thin even compared to aluminum foil. [According to wikipedia (the source of truthiness)](http://en.wikipedia.org/wiki/Aluminium_foil), aluminum foil typically ranges from 0.2 mm to 0.006 mm. Of course aluminum is stronger than lead.

**How much could their balloon have lifted?**

If they filled their balloon with pure helium (which they didn’t), how much would it lift? Well, there are essentially two forces acting on it. The buoyancy force and the weight of the stuff. In this case the stuff is the helium and the lead. (just as a side note: The helium doesn’t make it float. The purpose of the helium is to keep the walls of the balloon from collapsing. If you could make a material strong enough that it wouldn’t collapse (and be light enough) you could make it float with nothing inside). If used some other gas to fill it up (like argon), it would just add too much weight. For the Mythbuster’s balloon, the lead weighs 11 kg. There is 1000 cubic feet of helium (10x10x10). 1000 cubic feet is 28.3 m3. The density of helium (He) is 0.1786 kg/m3. So:

![Page 0 Blog Entry 14 9](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-9.jpg)

This would make a weight (force) of:

I also need to include the weight of the lead.

![Page 0 Blog Entry 14 11](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-11.jpg)

And now, the buoyancy force: (the density of air is 1.3 kg/m3)

![Page 0 Blog Entry 14 12](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-12.jpg)

Compare this to the claim from the Mythbusters that it would have 30 kg of lift (361 Newtons on the surface of the Earth could be the weight of 36 kg – of course I rounded in some areas). Thus the MBs (Mythbusters) were only talking about the lift of the shape, not the amount the object could lift. The total force on this lead balloon would be:

![Page 0 Blog Entry 14 13](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-13.jpg)

So, you could add another 45 lbs of weight and it would still float. This is assuming it was filled with helium (they used a mixture) AND that it was filled all the way (which they didn’t). The lead foil would probably tear if they filled it all the way up.

**How small could they have made the balloon?**

Clearly their balloon was huge. Their first attempt at a balloon was much smaller, but it did not float. The Mythbusters showed a quick picture of why they had to make it bigger. Basically, the weight of the lead is proportional to the surface area (since it is a constant thickness). The buoyancy force is proportional to the volume. So, if you make a cube twice as wide, what happens? Here is a generic cube:

![Page 0 Blog Entry 14 14](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-14.jpg)

This cube has sides of length d. The volume of this cube will be V = (d)(d)(d)= d3. The surface area of this cube (a cube has 6 sides) is SA=6*(d)(d) = 6d2. So, if I look at the ration of Volume to Surface area, I have:

![Page 0 Blog Entry 14 15](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-15.jpg)

The key point is that if I double the length of the side of the cube, I increase the volume (and lift) by a factor of (2)(2)(2) =8. I increase the mass of the lead by (2)(2) = 4. So, I gain lifting ability. (well, the balloon does)

**What would be the smallest size ballon (cube) that one could make with that thickness foil and have it float?**
Let me start with a cube of dimension (d) and calculate the the lift. The point is to make the net force (weight of helium, plus weight of lead plus buoyancy force) equal to zero. Here is the weight of the lead:

![Page 0 Blog Entry 14 16](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-16.jpg)

Note that the volume if 6d2t where t is the thickness of the foil.
And the weight of the helium:

![Page 0 Blog Entry 14 17](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-17.jpg)

And the buoyancy force:

![Page 0 Blog Entry 14 18](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-18.jpg)

This makes the total force (remember the buoyancy is pushing up and the two weights are pushing down:

![Page 0 Blog Entry 14 19](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-19.jpg)

Now, I simply need to set this total force to zero Newtons and solve for d:

![Page 0 Blog Entry 14 20](http://blog.dotphys.net/wp-content/uploads/2008/09/page-0-blog-entry-14-20.jpg)

I neglected to take into account the mass of the tape to hold the foil sheets together. So, if the mythbusters made a balloon square that was 1 meter on each side, it should float.
Of course the ginormous balloon they built was totally awesome and what makes the mythbuster the mythbusters. My hats off to you, Adam and Jamie.

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