Payload fraction as function of Δv and Isp: Foundations

This page assumes you have seen nothing. We will build every letter, ratio, and symbol the parent note Payload Fraction leans on, one at a time, each with a picture. By the end, every symbol in the master formula will feel obvious.

1. What a rocket actually is (a mass that splits in two)

Before any symbol, picture a rocket as a block of mass that is about to divide itself into two piles: a pile it throws away (burnt fuel shooting out the back) and a pile it keeps (the empty shell plus its cargo).

Figure — Payload fraction as function of Δv and Isp

The whole topic is bookkeeping on this one bar. So we need names for its slices.

2. The three slices of the bar

Look at the bar in the figure. It is cut into three labelled pieces:

Why these three and not more? Because after a burn, only two things can happen to a kilogram: either it left (propellant) or it stayed (payload + structure). No fourth category exists.

3. Before and after: $m_{0}$ and $m_{f}$

A burn is a movie with a start frame and an end frame. We need a symbol for each frame's total mass.

The subscript $0$ means "at time zero" (the start); $f$ means "final". The picture shows the tall bar ( $m_{0}$ ) shrinking to a short bar ( $m_{f}$ ) as the fuel pile disappears.

4. Ratios: turning "how much" into "how efficient"

Physicists rarely care about raw kilograms — a big rocket and a small rocket can be equally good. What matters is fractions: what portion of the whole is useful? A ratio is just one bar divided by another; it has no units, it's a plain number. Some ratios are forced to lie between 0 and 1 (a part of a whole, like the payload fraction below); others can be larger than 1 (comparing a big thing to a small thing, like the mass ratio below). Always ask "which is on top?" before assuming a range.

5. Why speed enters: $Δ v$

To go to orbit you must change your velocity. The symbol for that change is $Δ v$ .

Why is $Δ v$ the "cost" of a mission and not distance or time? Because in the near-vacuum of space there's no friction — what a rocket spends to reach a destination is velocity change, not miles. $Δ v$ is the fuel-budget currency. (When you fire deep in a gravity well you get bonus energy — that's the Oberth Effect — but the accounting unit is still $Δ v$ .)

6. Engine quality: $g_{0}$ , $v_{e}$ , and $I_{s p}$

Two engines can both burn fuel, but one throws it out faster. Faster exhaust = more push per kg of fuel = a better engine.

7. The tool that ties them together: the natural log and $e$

The parent formula contains $ln$ and $e$ . These scare people, so we build them from zero.

Where the $ln$ actually comes from

Let's watch the rocket lose one tiny bit of mass and see the $ln$ appear on its own.

WHAT we do: track the rocket at some instant with mass $m$ moving at speed $v$ . In a tiny slice of time it throws out a tiny mass $d m_{ex}$ of gas at exhaust speed $v_{e}$ (backwards). Conservation of momentum says the push the rocket gains equals the momentum carried away by that gas:

$m d v = v_{e} (- d m)$

Here $d m$ is the (negative) change in the rocket's own mass — it loses mass, so $- d m$ is the positive lump ejected. WHY this equation: it's just "momentum in = momentum out" for one tiny puff.

WHAT we do next: separate the two variables so each side has only one letter, then add up (integrate) every tiny puff from the full state to the empty state:

$d v = - v_{e} \frac{d m}{m} ⟹ \int_{0}^{Δ v} d v = - v_{e} \int_{m_{0}}^{m_{f}} \frac{d m}{m}$

WHY divide by $m$ : because each puff pushes a rocket that is lighter than the last time — the effect of a puff depends on the current mass, so $d m / m$ (fractional mass loss) is the natural quantity. Adding up many $d m / m$ pieces is exactly what produces a logarithm.

WHAT the integral gives: the running total of $d m / m$ is $ln m$ , so

$Δ v = - v_{e} [ln m]_{m_{0}}^{m_{f}} = - v_{e} (ln m_{f} - ln m_{0}) = v_{e} ln (\frac{m _{0}}{m _{f}}) .$

Swapping $v_{e} = I_{s p} g_{0}$ gives the form the parent uses:

So the $ln$ is not decoration — it is the accumulated cost of pushing an ever-lighter rocket, one puff at a time.

8. When one rocket isn't enough: staging

If one bar can't do the job (you'll see it go negative in the parent's Example 1), you drop dead weight partway and continue lighter.

How the foundations chain together

Read this as a dependency ladder — each rung needs the rung below it before it makes sense:

$λ$ (payload fraction, the goal) needs → $m_{0}$ , $m_{payload}$ , and $ϵ$ .
$ϵ$ needs → $m_{structure}$ and $m_{propellant}$ (the slices).
The mass ratio $m_{0} / m_{f}$ needs → $m_{0}$ , $m_{f}$ , and links to $ζ$ via $m_{0} / m_{f} = 1/ (1 - ζ)$ .
The Tsiolkovsky equation needs → the mass ratio, $Δ v$ , $I_{s p}$ , $g_{0}$ , and the $ln$ / $e$ pair.
Solving Tsiolkovsky together with $ϵ$ produces → the master payload-fraction formula.
Staging is the escape hatch when a single $λ$ comes out negative.

Equipment checklist

Test yourself — cover the right side and answer:

What does

m_{0}

mean, in words?

The total mass at ignition = payload + structure + propellant.

What is

m_{0} - m_{f}

physically?

The mass of propellant burned and thrown out.

Write the definition of payload fraction

λ

λ = m_{payload} / m_{0}

Write the propellant mass fraction

ζ

and its link to the mass ratio.

ζ = m_{propellant} / m_{0}

, and

m_{0} / m_{f} = 1/ (1 - ζ)

What does the structural coefficient

ϵ

measure?

Structure mass per kg of propellant — the packaging tax.

Why is the mass ratio always greater than 1?

A rocket can only lose mass, so the start is always heavier than the end.

What does

Δ v

measure and in what units?

The total velocity change a mission needs, in m/s.

How do

I_{s p}

and

v_{e}

relate?

v_{e} = I_{s p} g_{0}

, with

g_{0} = 9.81

m/s².

Where does the

ln

in Tsiolkovsky come from?

From integrating

d m / m

— the fractional mass loss of an ever-lighter rocket.

What problem does staging solve?

It sheds empty structure mid-flight so you stop wasting fuel carrying dead weight.