How is the efficiency of a rocket launch calculated when the weight keeps changing with fuel being spent + stages detaching?

Question

I assume it has to be a really complex calculation, how is the weight accounted for when the target number keeps changing constantly so dramatically?

In: 5

Answer 1

Staging is conceptually fairly simple – you take a small rocket with a payload and use a big rocket (booster stage) to lift the small rocket up and make it go fast. From a calculation perspective, you do the booster stage carrying the second stage and then the second stage by itself.

There’s a shortcut for figuring out the efficiency of a single stage known as the rocket equation. It’s written as:

delta v = 9.8 * specific impulse * ln(starting mass / final mass)

delta v is a measure of the useful amount of work the rocket stage can do. It takes about 9400 meters / second of delta v to get into orbit.

Specific impulse is a bit like fuel economy in a car. Rockets work by throwing mass out the back and it’s better to throw a small amount of mass really fast than a large amount of mass more slowly, as you get more effect for a given amount of fuel. Specific impulse depends on the fuel you use – hydrogen is better because it has light atoms that therefore go faster. Specific impulse also depends on the engine design; some engines use all the fuel to go forward, some waste some of it.

The other term is the natural logarithm (ln) of the ratio between the stage full of fuel and the stage empty of fuel. It’s basically a measure of how good you are at stuff fuel into a stage; a stage where 90% of the mass is fuel will be better than one where 80% of the mass is fuel. To do well here you want light tanks and light engines.

There’s tension between specific impulse and the fuel mass. Many people look at the very high specific impulse of hydrogen and don’t notice that it’s really hard to fit hydrogen into tanks because it is not dense at all. That’s what NASA’s SLS has a core stage that is so huge.