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

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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

You use a differential equation – one that couples variables with their *rate of change* to determine that. Year 1 college physics level of complexity.

The simplest is to think of the useful energy output: the increase in gravitational potential energy and kinetic energy of the payload. The total input chemical energy is far higher since most of that energy is used to transport the fuel etc.

Since the input energy is dependent on the exhaust gas velocity and flow rate (as well as number of stages etc) efficiency is limited by mechanical design and chemistry.

The most basic form of the rocket is basically a tube that farts out burned fuel. As you have said the weight of the rocket is not constant, which might make the problem sound complicated, but dealing with changing quantities is the essence of CALCULUS.

with a bit of calculus tinkering, you can derive the “rocket equation”(google keyword), which basically allows you to calculate the change in speed with the fully fueled mass, empty mass and exhaust velocity

Not exactly 5 yo, but [here’s](https://www.youtube.com/watch?v=bPXjkFAnQio) how you can derive the “rocket equation”,

Let’s simplify to an an easier problem first. If you drive the car at constant speed, and you know the speed of the car and time you spent driving, how to you calculate the distance you traveled?

Easy. Distance = speed × time

But, what if you’re speeding up, i.e. your speed is changing just like the rocket weight is constantly changing?

Let’s say the [red line in this graph](http://calcprojectpo.weebly.com/uploads/2/9/7/4/29745541/2512833_orig.gif) is your speed graph. You can’t just multiply speed with time because speed is changing. But you can take smaller chunks of time (let’s say, 0.5 seconds) and make an approximation by saying that your speed was constant during that small chunk of time because that’s not enough time for your speed to significantly change. Those are the green rectangles in the image.

Now that you’ve taken a small chunk and assumed the speed is constant during that chunk, you can multiply distance = speed × time. So, your distance traveled during the first chunk would be, say, 5 km/h × 0.5 s; your distance traveled during the second chunk would be 5.2 km/h × 0.5 s, third chunk 5.4 km/h × 0.5 s. As you add up, say 100 chunks, you can know how far you’ve traveled during 50 seconds of your ride.

The smaller the chunks, the more accurate your result because your approximation more closely resembles the actual situation.

Similarly, a rocket scientist would take a mass of the rocket at a certain small chunk of time, assume that weight and speed are constant during that chunk of time, and make his calculations easier for that particular chunk. Then you add up all the chunks and you’ve got your result.

That’s called calculus. It’s basically a branch of math that, among other things, deals with quantities that change with other quantities (i.e. speed that changes with time, rocket acceleration that changes with mass, altitude, air resistance, …).

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.