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The art of estimations is pretty much a physicist's bread and butter. We all love a good estimation problem. You might also hear these called a "back of the envelope" calculation, or a calculation on a napkin. The writing medium is meant to emphasize how little preparation goes into attacking the problem. The estimator can't even take the time to find a clean sheet of paper.
But these calculations aren't just for fun, they are also useful. One good example is a question that I just recently calculated: How big of a balloon would you need to lift a human?
Woah - People - Wait - Air - Balloon
Woah! Right there people might be saying: But wait! Is this a hot air balloon? Is this a hydrogen balloon? What material is it made of? The answer is "I don't know, I don't care." The true power of the estimation problem is that you can pretty much ignore things that you don't know. Yes, this takes some practice to figure out what is important and what isn't. With that said, let me start off with my assumptions and estimations.
The buoyancy force from a balloon is equal to the gravitational weight of the air displaced.
Density - Air - Kilogram - Meter
The density of air is around 1 kilogram per cubic meter.
A human has a mass of 90 kilograms.
Balloon - Cube - Side - Length - L
A balloon is a cube with a side length of L (of course it's not really a cube).
I'm going to ignore the weight of the air and balloon material.
Order - Balloon - Buoyancy - Force - Weight
Let's begin. In order for a human to be lifted by a balloon, the buoyancy force must be equal to the weight of the human. The buoyancy force for any object is equal to the gravitational force of the air (or water) displaced by that object. If I know the density of air (I do), I can solve for the volume of the...
(Excerpt) Read more at: WIRED
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