About a century ago, astronautics pioneers such as Konstantin Tsiolkovsky developed the rocket equation. The rocket equation gives the final velocity of a rocket as a function of the fraction of its initial mass expelled as reaction mass. It is the ideal case, in a vacuum, in the absence of gravity..

Vf = Ve ln R

where

Vf = final velocity (of the rocket after the burn, the delta-vee)

Ve = Exhaust velocity (velocity of the exhaust gases relative to the rocket).

ln = the natural logarithm function (log to the base e)

R = the mass ratio, Mi/Mf, or the Initial Mass (propellant+rocket+payload) divided by the Final Mass (Rocket, payload, plus unburned propelllant, if any)

You will note this equation works for any propulsion system that expels reaction mass (a rocket), regardless of the type fuel used or the technical efficiency of the engine. The exhaust velocity Ve is a constant which multiplies the log of the mass ratio by a number representing the effectiveness of the fuel/engine. In other words, all else being equal, the higher the exhaust velocity, the faster you go. The mass ratio R tells you how much propellant was burned and expelled as reaction mass. The higher the ratio (the more you burn) the faster you go.

But as you may have guessed, there is a Law of Diminishing Returns at work here. True, as R gets bigger (the more fuel you burn), the faster you go, but doubling R doesn’t mean you go twice as fast. In fact, for relatively high values of R the additional speed you get by carrying more fuel gets very expensive. The following table will give you some idea how this works, where the empty rocket masses 1.0 (ton, pound, whatever) and the final velocity goes up as R goes up. Ln R is the multiplier with Ve that gives you final velocity. BTW, the Windows calculator has a natural log “ln” key so you can do your own calculations. Let’s assume a Final Mass of unity (1 ton, pound, kilo, whatever) for all cases So that R-1 = mass of propellant consumed in terms of the empty rocket mass. By empty mass, I mean the mass of the rocket airframe, empty fuel tanks, engines, crew, cargo, and any additional fuel required for any further maneuvers.

R…..ln R

1……0.00 (The rocket carries no fuel)
2……0.69 (The fuel mass = empty rocket mass)
2.72..1.00 (Fuel mass = 1.72x the rocket mass
3……1.10 (Fuel mass = 2x the rocket mass)
4……1.39 (Fuel mass = 3x the rocket mass)
4.92..1.59 (Fuel mass = 3.92x the rocket mass
5……1.61 (Fuel mass = 4x the rocket mass)
…
9……2.20 (Fuel mass = 8x the rocket mass)
10…..2.30 (Fuel mass = 9x the rocket mass)

Its pretty clear that by simply adding more fuel you can always go faster, but at high mass ratios you don’t get that much speed for the extra fuel you need to carry. You waste most of your fuel energy pushing unconsumed fuel!). And it becomes very difficult to build a rocket whose mass is mostly made up of propellant, considering the tankage, piping and pumps needed to contain and manage it. Rocket engineers deal with this problem by using multiple stages, so that empty fuel tanks and the machinery needed for it can be discarded to save weight as propellant is consumed.

This state of affairs is modeled by the natural logarithm (ln) function. We all know what logarithms are: In high school we all learned about logarithms to the base 10 (abbreviated “log”). So for example, the log of 10 is 1, the log of 100 is 2, the log of 1000 is 3, the log of a million is 6. These base-10 logs were invented as an aid to calculation centuries ago (the sum of the logs of two numbers equals the log of their product). Slide rules used logs to aid in calculation, today we use electronics so base 10 logs have fallen out of fashion. But in nature, the natural logarithm to the base e shows up in all sorts of unexpected places, kind of like pi. But just what is “e“, anyway? Taking the log is the opposite of exponentiation. The two operations are related,.

If y = e^x then x = ln y
or, in a base 10 example
If 1000 = 10^3 then 3 = log 1000

Bear with me, I’ll get back to rockets in a minute.

e, the base of the natural logarithms, like pi, has an actual value, it is a transcendental number which cannot be represented exactly by any fraction or any finite number of decimal places. Like pi, the first five places are good enough for most engineering calculations.

e = 2.71828…

e is one of those magic numbers, like pi, that shows up all over mathematics and physics. We know what pi is, the circumference of a circle divided by its diameter, but where does e come from? And what does it have to do with rockets?

You’ve no doubt heard of the term “exponential”, as in “exponential growth”. The phrase is misused and abused in common speech, but it has a strict and clear mathematical definition. It means a quantity which grows (or shrinks) at a rate related to the quantity there is available at the time. Mathematically this is expressed as the exponential function y = e^x. As x increases, y grows, explosively. Expressions of this form are useful for dealing with processes involving rates of change, such as the growth of bacterial populations, radioactive decay and the behavior of rockets. But where does e come from? Why is e chosen as the base of the exponential function?

The graph of y = e^x has a very unusual and useful property: The rate of growth of the curve at any point (think of it as the slope of the line at that point) is equal to the value of the curve at that point. In calculus terminology, the value of y=e^x is equal to its own derivative. e is the value that governs how things change in the universe.

Anyway, I digress. Of what possible use is all this nonsense?
Well, for starters, if we wanted to design a rocket that was efficient, that is, carry as much payload as possible while going as fast as possible (remember, the two work against each other, so we are looking for a compromise here, an optimum value) what mass ratio should we design for?

Well, the higher the mass ratio, the faster the rocket. Vf = Ve ln R. as we have seen, this function has no inflection points, Vf simply continues to increase for higher values of R, although
it increases less dramatically.

And the lower the mass ratio, the more dry weight (rocket+payload+unburned ) for a given total mass. The most efficient design R would be as high as possible for greater velocity, but as small as possible for greater cargo capacity, clearly contradictory requirements. We must find the optimum combination of these two competing requirements.

So the efficiency ratio Vf / R must be maximized, or

Efficiency = Ve (ln R) / R be made as big as possible.

And it just turns out that the value of R that gives you the best efficiency, that is, the optimum ratio of maximum final velocity to maximum dry rocket mass is 2.71828…, the value of e! About 63% of the ship’s total mass is fuel. Of course, the “dry rocket mass” might include the propellant you need to brake when you get to your destination. For a hundred ton rocket, you would require 172 tons of propellant and it would get you up to 100% of Ve. You could go faster by adding more fuel, but you would have to sacrifice too much payload capacity.

That’s the calculation you’d make for a cargo rocket, where you want to go as fast as you can AND haul as much payload as possible. If simply going as fast as possible with as little propellant mass as possible was all that mattered (like say, a torpedo!) then the required mass ratio would be about 4.92, and your final Vf would be 1.59 Ve. This mass ratio means about 80% of your spacecraft is propellant mass, and it extracts the most possible potential energy in the propellant and converts it to kinetic energy of dry (empty) rocket. Getting your unburned fuel up to speed requires energy, but rocket kinetic energy goes up as the square of Vf, the optimum tradeoff is at R = 4.92. Again, you can go faster by carrying more fuel, but the weight penalty is enormous.

To summarize, your hundred ton dry rocket must carry 172 tons of propellant to be an efficient cargo hauler. If speed only is the object, 392 tons of fuel is the optimum amount. In spite of adding an extra 220 tons of propellant, more than doubling your fuel load, your speed only goes up by 59%. If you are captain of a warship and need to carry as many fast torpedoes as possible, you will opt for an R of 2.718, but your missiles would have an R of 4.92. If you tried for faster torpedoes, or a faster ship, you’d have to pay the price by carrying fewer torpedoes.