Pretty much everyone loves balloons—especially younger children. Kids are slowly building up ideas about the way the universe works (through their observations), and they already know that when you let go of something, it falls. Oh, but the helium-filled balloon is a rule breaker. It goes UP. It just seems magical.

Older people still have a hidden fascination with these balloons. Each of us at some point has considered the question: How many of these would I need to lift me off the ground? Well, that's exactly what David Blaine did for his latest stunt, which he called Ascension. He used a bunch of large balloons to lift him up to an altitude of 24,000 feet. At that point, he detached himself from the balloons and used a parachute to get back down.

I think the best part of the stunt was the initial launch. The team set up the balloons so that there was a near perfect balance between the buoyancy force from the balloons and the gravitational force pulling Blaine down, such that he mostly just floated there right above the ground. (He did have some people holding on to him to make sure he didn't drift up and away prematurely.) Then, so he could begin his journey upward, his daughter added one more balloon, and he handed her a weight he'd been holding. It's a pretty cool way to ascend.

But now for the questions and answers.

Why do helium balloons float?

Balloons don't float with magic. Instead, it's a result of gravity and the atmosphere. Yes, it's true. A balloon wouldn't float without gravity.

Let's imagine the atmosphere as a bunch of balls—except these balls are actually molecules of mostly nitrogen along with some oxygen. Each of these balls are moving around with some average velocity, and they are being pulled down by the gravitational interaction with the earth. So, you could think of these gas balls just like a tennis ball tossed across the room, except that they are super tiny. Oh, and there's a bunch of these balls. That means that they interact with other gas balls. You can think of these interactions as though they were collisions. It’s all of these ball-ball collisions that keeps them from just ending up on the ground. Also it would be terribly awkward if all the air pooled down on the lowest level, because then you couldn’t breathe.

When two gas balls collide, sometimes one of the balls gets deflected upward, and sometimes it gets deflected sideways. However, since there is also a gravitational interaction pulling the balls down, there are more of them closer to the ground. This is why the density of air decreases as you move vertically upward. The density of air near the ground is about 1.2 kg/m^{3} and decreases to around 0.59 kg/m^{3} at an altitude of 7,000 meters (close to 24,000 feet). But even over a distance from the bottom of a balloon to the top, the density of air changes—just a little bit.

Now let's put an object in the air. I'm going to use a brick. I like the brick since it clearly doesn't float in the air, but also it has flat surfaces to make my explanation easier. Since the tiny air balls are moving around, some of them are going to collide with the surface of the brick. When a ball bounces off the brick, it gives a tiny little push on that brick. The total force on one surface of the brick depends on the area of this brick and the pressure of the air. Just a reminder, the relationship between force and pressure can be expressed as the following equation, where *P* is the pressure, *A* is the area, and *F* is the force.

So, if you have a large surface area and a small pressure, you can still get a large force. In this expression, the pressure is due to the atmosphere—that’s those gas balls moving around and colliding with stuff. Here’s the cool part. Because there are more gas balls closer to the ground, the pressure depends on the density of air, and, remember, the density depends on the altitude. This means that the force from the air pushing on the top of the brick is different than the force on the bottom of the brick. It's best to describe these collisions in terms of pressure and model the change in pressure with the following equation.

In this expression, P_{0} is the pressure at some arbitrary point where y = 0 (in the vertical direction), *g* is the gravitational field (9.8 N/kg) and ρ is the density of air. So as y increases, the pressure decreases. Note: This linear relationship is only approximately true. When you get really far above the surface of the earth, it doesn't work. But with this, you can see that the force from the air on the top of the brick should be less than the force on the bottom of the brick.

Notice that the forces pushing on the left and right sides of the brick are at the same height. This means that the net force in the horizontal direction would be zero—they cancel. But the force pushing UP on the brick (from the bottom) is greater than the force pushing DOWN since the bottom of the brick is at a lower altitude—even by just a little bit. If the brick has a height *h*, then the total force from the air in the vertical direction would be:

Notice that I skipped some algebraic steps, but it's not too difficult to see how that works out. But wait! If I multiply the height of the brick (*h*) by the area of the bottom (*A*), I get the volume (*V*) of the brick. Then, if I multiply the volume of the brick by the density of air (ρ), I get a mass—the mass of area with the same volume as the brick. When you multiply that mass and the gravitational field (*g*), you get the weight of the air displaced by the brick.

Boom. This is the famous Archimedes' principle. It says that when an object is in water, there is an upward buoyancy force on the object. The value of this buoyancy force is equal to the weight of the water displaced. But it also works for displaced air. Yes, there is an upward buoyancy force on the brick. The brick doesn't float like a balloon because there is also a downward gravitational force on the brick—and this downward force is much greater than the upward buoyancy.

Oh, here is the cool part. It doesn't even matter if you replace the rectangular brick with a spherical balloon. The buoyancy force still just depends on the density of the air and the volume of the object. So, why does a helium balloon float? The only thing special about a helium gas is that it has a significantly lower density than air (with a density of 0.179 kg/m^{3} for helium and 1.2 kg/m^{3} for air). This means that the gravitational force pulling down on the balloon would be smaller than the upward buoyancy force, and it would float. Just to be clear, a water-filled balloon and a helium balloon of the same size have the same buoyancy force. It's just that the weight of the water-filled balloon is huge.

How many balloons do you need to lift a person?

I'm not saying you should float yourself up into the air with a bunch of balloons, but let's say you want to estimate the number of balloons you would need. It wouldn't be too difficult to calculate the volume of air that would have a weight equal to the weight of a human and then find the volume of helium you would need, but that neglects something very important—the rubber in the balloon. Yes, it has a small mass, but it still matters. Let's say that I have some generic spherical balloon made of rubber of some arbitrary thickness. Maybe it looks like this.

This balloon has a radius *R* with a rubber thickness *t*, and it's filled with helium. I need to find the mass (and therefore the weight) of both the helium gas and the rubber. Let me call the density of helium ρ_{h} and the density of rubber ρ_{r}. The weight of the helium depends on the volume of the balloon. Since it's a sphere, the weight of the helium would be:

Yes, I used the volume of a sphere in there. Now for the weight of the rubber. I need the volume of this thin shell on the outside of the balloon. If the thickness of the rubber is small compared to the radius of the balloon (which is approximately true), then I can calculate the rubber volume as the surface area of the sphere multiplied by the thickness. This gives a rubber weight of:

There is that parameter *t* in the weight of the rubber. Here's the deal, you can't make this as thin as you like. There is some limit—so let's just say it's a constant value. That means the rubber weight is proportional to the square of the balloon radius, but the weight of the helium is proportional to the CUBE of the radius. The helium has a much lower density than rubber, so you want a large helium-to-rubber ratio, and that means bigger balloons are better.

If you take your standard party balloon, it has a fairly small radius (let's say 10 cm) such that you waste a lot of mass on the rubber. However, if you get a much bigger balloon like in Blaine's Ascension stunt you will get a much better helium-to-rubber ratio.

OK, now for a rough estimation. I'm just estimating stuff here—because that's what I do. I'll start with a rubber density of 1,000 kg/m^{3} which is the same as water (close enough to rubber). For the balloon radius, I will use 0.75 meters and a thickness of 0.2 mm. That means the net lifting force for one balloon would be:

I know that looks crazy, but it's not. It's just the weight of the displaced air minus the weight of the helium and rubber. Now to find the number of balloons, I just take the weight of the person (let's use David Blaine plus other equipment with a mass of 100 kg) and divide by the lifting force for one balloon. Here is the calculation as a python script (so you can change the values).

Oh, that's not good. 256 balloons is not going to look epic for a YouTube show. Of course, I could be totally off on my estimation of balloon thickness—but check out what happens if I change the radius to 1.5 meters. I get about 11 balloons. That seems better. Quick note: That calculation above is actual code. If you click the pencil icon, you can see my estimated values and change it to whatever you like. Then click the Play button and run it.

Would the balloon keep rising forever?

Obviously nothing goes on forever. A balloon will keep increasing in altitude as long as the lifting force is greater than or equal to the overall gravitational force pulling down. The thing that is going to change is the lifting force. At higher altitudes, the density of air decreases. This means that since the buoyancy force is equal to the weight of the air displaced, it will also decrease.

So, the balloon will eventually reach an altitude that puts it in equilibrium, and it won't go any higher. Of course this assumes the volume of the balloon also remains constant—which isn't technically true. At high altitude, the atmospheric pressure decreases and pushes less on the balloon. This means that the helium inside the balloon can stretch the rubber and expand and produce more buoyancy force. It's also that at some point, the rubber will stretch too much and then break. This would be bad, since all the helium would escape and you would just have a big piece of rubber. That's not very helpful.

What is the acceleration on takeoff?

I want to get an estimation of his vertical acceleration at the beginning of the ascent. There's not a perfect camera angle, but I can roughly estimate his position in different frames of the video (to get the time). With that, I get the following plot of vertical position as a function of time.

If an object has a constant acceleration, its position can be found with the following kinematic equation.

The important thing here is that I can use this equation to find the value of the vertical acceleration. If I fit a quadratic equation to the data, the coefficient in front of the t^{2} must be equal to the *(½)a* term in this kinematic equation. That means I can use the fit to find the acceleration, and I get a value of about 0.05 m/s^{2}. Yes, I skipped some steps here, but you can fill in the missing parts as a homework assignment. But is this value even that reasonable?

How about we approach this with another method? Let's say Blaine is in equilibrium with a net force of zero newtons. He then hands a small weight of 1 pound to his daughter (4.4 newtons). Oh, there is also that extra balloon his daughter added. But I think for this estimation we can just consider the handed weight. That means his weight decreased by 4.4 newtons to give a net upward force of 4.4 newtons. Now, I can use Newton's second law that says:

For the mass, I need the mass of both Blaine AND the balloons. Let's say this is 110 kg. With a force of 4.4 Newtons, the vertical acceleration would be 0.04 m/s^{2}. OK, that's actually closer than I thought it would be. I'm going to call it a win.

David Blaine successfully got his balloon rig up to an altitude of over 24,000 feet AND he parachuted back to the ground. I'm sure we can all agree that also is a win.