With an air density of 1 kg/m3 and a human mass of around 90 kg, the volume of the balloon would have to be 90 cubic meters. Now I just need to take the cubed root of 90—without a calculator because that would spoil the process. OK, cube roots are tough, so let's just get a rough value. I know that 5 cubed is 125 and 4 cubed is 64. I'll just split the difference and say that the cubed root of 90 is 4.5. That means the cube is 4.5 meters across or about 12 feet. The balloon would have a diameter of about 12 feet.

Halie Chavez

OK, this answer is wrong. It's wrong for lots of reasons. I didn't use the correct shape of the balloon and I used the wrong value for the density of air (it's really closer to 1.2 kg/m3). The biggest mistake is that I didn't take into account the mass of the gas in the balloon and I didn't include the mass of the balloon material. But being absolutely right isn't the point of an estimation. With my estimation, I at least have a rough answer. I know that this is going to have to be bigger than the biggest party balloon you can find at a store. You aren't going to be able to even grab a bunch of balloons from a performing clown and use that to float away. It's got to be a BIG balloon. Bigger than 12 meters across.

See? Wasn't that estimation fun? I'll do another one, but let me first go over some of the key components of a great estimation.

You have to start with something. There has to be some estimate or physics principle that you start with. You can't just estimate the mass of the moon if you don't know anything about gravity or how long it takes for the moon to orbit the Earth. In the example above, I just happened to know the density of air because I've used it before. I also knew the relationship between buoyancy and the size of an object (Archimedes' Principle). Without that, I would be pretty much stuck.

Approximate values are fine. Does it matter that I used an air density of 1.0 instead of 1.2 kg/m3? Not really. It's true that an air density of 1.2 kg/m3 would give a smaller size of the balloon—but not significantly smaller. Remember that I'm not trying to calculate the exact size of the balloon. I just want to see if it's a reasonable size. So, go ahead. Use a value of 10 N/kg for the gravitational field (instead of 9.8 N/kg). Use a human mass of 100 kg if that makes things easier. In the end, it won't matter too much.

Estimation problems are sort of like a rough pencil sketch. It doesn't have all the exact details of a photograph, but it's enough to give you an idea of what's going on.

Pick the simplest version of the problem. If you want to estimate the thermal radiation from a cow, you need to know the surface area of said cow. That's complicated, so let's make it simpler. What about a spherical cow? (classic physics joke) Even though cows aren't actually spherical, they are close enough to being spherical for this problem. Also, the surface area of a sphere is pretty easy to calculate.

If you don't know a value, at least get a reasonable range. But wait! What if I don't know the density of air? How could I do the balloon problem? OK, let's get started. What is some density value that you DO know? What about water? That has a density of 1000 kg/m3. Should air have a higher or lower density? Yes, lower. But how much lower? 100 kg/m33 seems too high still. What if you used 10 kg/m3? That's pretty high—but it's at least a start. Also, the density is clearly greater than 0 kg/m3. How about a range from 0.1-10 kg/m3. See, at least we have something to work with now.