Dot Physics

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)
http://phet.colorado.edu/new/simulations/sims.php?sim=Balloons_and_Buoyancy

i-aed22c4d5cb001e39b56c3c727e0708b-2009-12-24_phet.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:

i-15f9c3a2e5aee3e8e59599575ba6b301-2009-12-24_gassforces.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:

i-39a086e579e1c5d34d1793b9638be4cb-2009-12-24_la_te_xi_t_1.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:

i-133608a84e2939dc3bf4a5cc0eda52a9-2009-12-24_area_1.jpg

Such that it has a volume of:

i-93db4fb97011cd377006ec136c0eeee0-2009-12-24_la_te_xi_t_1_1.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:

i-ae5298c3b2253b1f29e8834f83456025-2009-12-24_la_te_xi_t_1_2.jpg
and
i-2189513131d9dace722acbe4d498b91b-2009-12-24_la_te_xi_t_1_3.jpg

This would mean that the thickness would be:

i-27966b5569245ef5f32856042758a7f3-2009-12-24_la_te_xi_t_1_4.jpg

That’s pretty thin. This is thin even compared to aluminum foil. [According to wikipedia the source of truthiness, 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:

i-61fb786142d08abc74282d49f0f489e7-2009-12-24_la_te_xi_t_1_5.jpg

This would make a weight (force) of:

i-b90c0fc26879719f4976375822f38859-2009-12-24_la_te_xi_t_1_6.jpg

I also need to include the weight of the lead.

i-d447d95a70d6251ce10af4627e1b339e-2009-12-24_la_te_xi_t_1_7.jpg

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

i-55797076efe860803e2531a4b8975fe7-2009-12-24_la_te_xi_t_1_8.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:

i-e5fc8a0c8d5c963105158dbfda22d60e-2009-12-24_la_te_xi_t_1_9.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:

i-87165e7f6c0ae4175ad1bc6b6040671c-2009-12-24_area_2.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:

i-e24b0f6ea608a2c18d8560e2d0bb05bd-2009-12-24_la_te_xi_t_1_10.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:

i-57f2f35a25d90d9786f733cb2f098196-2009-12-24_la_te_xi_t_1_11.jpg

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

i-ef3eb84b4cfb0317355af8322de49614-2009-12-24_la_te_xi_t_1_12.jpg

And the buoyancy force:

i-cd1cac6b85db1a4d55c85ab5c16a7b69-2009-12-24_la_te_xi_t_1_13.jpg

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

i-cb16f97721c2caba1518541ddeb45a4f-2009-12-24_la_te_xi_t_1_14.jpg

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

i-6b360a95d1a7aa0ce7fed9b52f8e300c-2009-12-24_la_te_xi_t_1_15.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.

Comments

  1. #1 _Arthur
    December 26, 2009

    Can’t we just use simple Archimedes ? If the weight of the lead enclosure + the weight of the gas inside is LESS than the weight of the same volume of air, then it will rise ?

    The weight/mass of the gas is determined by the volume and the pressure.
    We can assume that the minimal pressure to properly inflate the balloon will be the same as ambient air pressure (very slighly higher being more practical).

  2. #2 _Arthur
    December 26, 2009

    addendum: the mass of the ambient air, and of the helium, also depends of their temperature. In their stunt, the helium is likely to have been (initially) cooler than the air temperature, being just depressurized from a high pressure tank.
    I don’t remember if the mythbusters attempted to mitigate that factor.
    Cooler than room temp helium would be more dense, hence weight more for its volume.
    Heating the helium to more than room temp would have violated the spirit of the experiment.

  3. #3 Cleon Teunissen
    December 26, 2009

    Rhett wrote:
    “First, how does stuff float at all? There are many levels that this question could be answered.”

    Remarkably, it’s not immediately clear which level is best. Rhett builds up the concept of buoyancy from first principles. Commentator _Arthur wonders whether it would have sufficed to invoke Archimedes’s principle. I go with Rhett.

    Archimedes’ principle deals with the question whether a particular object will be buoyant or not. If you can find out the density of an object in advance you can predict whether or not it will float. But the deeper question is the physics of buoyancy itself.

    In the vast majority of classroom physics demonstrations any pressure gradient in the room is ignored; it’s negligably small, just as gradient in gravity from floor to ceiling is negligable. But hang on – balloon buoyancy is an exception to that. A helium filled balloon that is a foot in diameter has enough buoyancy to float up. The pressure gradient of the atmosphere within a foot of height is the source of that balloon’s buoyancy.

    (In air we have that pressure gradient and density gradient will be pretty much proportional. However, in liquid there is a pressure gradient, but hardly any density gradient. Hence I prefer to emphasize pressure gradient as the source of buoyancy.)

    Archimedes’ principle offers economy of thought; Archimedes’ principle allows us to think about buoyancy in an abstracted form, it makes thinking about buoyancy more efficient. But this abstraction also has a hiding effect, it hides the underlying layer from view.

    So I think it really good to always try and probe deeper, to try and dig down to first principles, rather than relying on abstractions.

  4. #4 Rhett Allain
    December 26, 2009

    @_Arthur,

    I agree with Cleon. As physicists, we try to break things down into the most fundamental ideas. If we didn’t do this, we would be called chemists (ha!)

    Sorry chemists, it was just a joke.

  5. #5 Lord Axil
    December 26, 2009

    Ok, here’s a question to show the utility of above explanation of buoyancy.

    If you’re driving a car at constant speed in which a balloon is floating, initially with neutral buoyancy, and you step gently on the brakes, which way does the balloon move relative to an observer inside the car?

  6. #6 CCPhysicist
    January 16, 2010

    _Arthur, he did use simple Archimedes … for the CALCULATION.

    The explanation and the final equation are rarely the same thing.

    What I found most interesting in your calculation is that the mass of the He inside their balloon was nearly half of the mass of the Pb used to make the balloon. How many of your students would guess that?

    One thing I do as an example in class is calculate the mass of air in our classroom. The answer blows their minds. They have never thought about whether part of the mass of the space station is the mass of the air inside because people tend to think of air as nothing.

  7. #7 Just Some Guy
    June 25, 2010

    Umm, shouldnt we be using hydrogen instead of helium to fill the balloon?

    I remember the Hindenburg too, but in principle, to make the smallest floatable lead balloon, we gotta use the lightest gas to fill it, right?

    OK but we gotta use enough pressure to keep the ballon from collapsing, and we gotta recognize that lead is kinda soft, and since H atoms are little some of ‘em are gonna escape from a thin balloon of lead, so how much H pressure do we need in what size Pb object, assuming STP, for how long, to adequately demonstrate the mininum of plumbumational aerational flotational potentialities?

    Doeas anyone here have a well-reasoned estimate of the smallest size of a floatable in air (from sea level on Earth) lead ballon?

    I’d love to be able to tell people the size of the smallest floatable lead ballon, but darn it, I suspect that bringing up the concept, without knowing the empirically and properly scientifically established facts, would probably kinda go over like a lead ballon.

  8. #8 Henry
    January 1, 2011

    Rhett i have an amazing idea involving thease equations that could change the world, please email me at henry562@live.com so we can talk.

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