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A recent XKCD What-if article mentions the situation where each additional kilogram of cargo to LEO requires an additional 1. 3 kilograms of fuel, which in turn requires fuel to carry (simplification). I wonder then, if an additional half-kilo object, such as an iPod will have a large impact on the amount of fuel that engineers will load into the ET. Is the ET, and other rockets, always 'topped off' regardless of payload, or is the amount of fuel loaded carefully measured? What tolerances are acceptable? How much excess fuel is typically in the rocket / tank when the payload is separated?

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3 Answers 3

Consider Tsiolkovsky's rocket equation

$ \Delta v = v_e \ln \left( m_0/m_f \right) $

with $\Delta V$ the total change in velocity, $v_e$ the exhaust speed of the reaction products, $m_0$ the initial mass (structure+payload+propellant) and $m_f$ the final mass (structure+payload).

If you ignore the atmosphere and other such "nuisances", it should be obvious that for any given rocket $\Delta V$ is a value independent of the rocket's mass -- the rocket needs to be given a certain $\Delta V$ to reach space. Also, $v_e$ is a value that depends mostly on engine specifics, and is thus fixed for a specific type of rocket. It follows that $\Delta V = C_1$, a constant, and $v_e = C_2$, another constant.

You can then rearrange Tsiolkovsky's rocket equation to

$ m_f = m_0 e^{-\Delta v/v_e} = m_0 e^{-C} $

where simply $C = C_1/C_2$. Now, if some extra mass $a$ were to be added to the rocket's payload, this would translate into

$ m_f + a = a + m_o e^{-C} = \left( a e^{C} + m_0\right)e^{-C} $

in other words, it would require an additional $a e^{C}$ kilograms of propellant (note that I'm avoiding the word fuel, as that is generally used to describe a substance that reacts with oxygen, which isn't necessarily the case for rockets). The term $e^{C}$ is commonly called the growth factor.

As a numerical example, suppose

$ \matrix{\Delta V = 15\ \mathrm{km/s}\\ v_e = 7\ \mathrm{km/s}} $

which is roughly the maximum capability of today's technology (and yes, ion engines are much better in this respect, but these can not be used for manned launches, so we'll not consider them here). Then

$ e^C = e^{15/7}\approx 8.5 $

so you'll need approximately 8.5 kg extra propellant for each added kilogram of payload.

"Topping off" the rocket as you suggested can be rephrased as adding propellant mass to payload mass, which has possibly severe penalties on the maximum deliverable payload mass. Although a safely factor of around 1.1 is usually employed, rocket fuel is expensive, so "topping it off" is usually not done.

Moreover: engines running on solid propellants (as in booster rockets) can in general not be turned off, which would make topping off a complete waste as boosters are usually discarded to save on final structural mass. Also, many non-booster rocket stages are discarded mid-flight for the same reason, so topping off would only ever make sense in the final stage, provided that that stage is part of the payload and therefore not discarded.

So in summary: aside from a small safety factor, launch vehicles are filled with an amount of fuel specifically tailored to the mission.

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What about fission rockets? –  huseyin tugrul buyukisik Aug 17 '12 at 13:25
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Those do not exist yet (and I doubt they ever will due to political considerations). Nevertheless, for any type of rocket engine that uses reaction mass as its sole source of thrust, these equations apply. Obviously, if you'd increase $v_e$ while preserving large thrust, you're far better off :) –  Rody Oldenhuis Aug 17 '12 at 13:30
    
Thank you. You briefly mention a safety factor of 1.1. Does that mean that roughly 9% of the launch fuel remains? –  dotancohen Aug 17 '12 at 14:26
    
Yes, 10% is indeed included in launch planning. Mind you, it is only a rule of thumb: numbers like 2% or even 0.5% are also not uncommon. –  Rody Oldenhuis Aug 17 '12 at 14:28
    
Euhm, I responded too quickly. That is: $1.1 m_0$ is used, so $0.1 m_0 e^{-C}$ would remain in the final stage –  Rody Oldenhuis Aug 17 '12 at 14:30
What tolerances are acceptable?

If I were the man that calculates this, I would calculate the computer error range. Then human-error range. Then think of the worst-case (maximum error). Then multiply that error-fraction with the total amount of fuel. Then add that much fuel onto the existing fuel.

Errors would come from:

  • Measurement (both human and devices)
  • Calculations(both)
  • Probability of fuel leak.
  • Probability of sudden changes in atmosphere.
  • probability of finding free funds (credits or money whatever it is)

Lets say worst case is 0.5%, I'd put 0.5% more.

Lets assume all computers and all the silicon-germanium was stolen by aliens. What do we do without digital devices? We make it hand-calcuated. Lets assume this error is 40%. Then we put an extra fuel tank to be sure.

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Or simply learn to calculate complex things by hand again, like they often did during the Apollo moon landing. –  Rody Oldenhuis Aug 17 '12 at 13:32
    
Even calculating a square root with Taylor Series would take some minutes :P not even 1 FLOPS –  huseyin tugrul buyukisik Aug 17 '12 at 13:37
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Why would you do that if you can do this? –  Rody Oldenhuis Aug 17 '12 at 13:38
    
Shifting n-th root algorithm has cool precision with a few computing –  huseyin tugrul buyukisik Aug 17 '12 at 13:45
    
Usually, roots-by-hand are done with logarithms. everything beyond adding/subtracting done-by-hand is done with logarithms. Memorizing the logarithms of all integers from 0 to 100 is one of the most useful things you can waste your time on :) –  Rody Oldenhuis Aug 17 '12 at 13:47

See How Apollo flew to the moon the first lander and LCM had a large fuel reserve because they weren't sure and this was reduced to the final mission which returned to earth orbit with some very small amount left. Sorry don't have the book here for the actual figures.

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