Chapter 3. When Do We Need a Lander?

This chapter is less about action and more about calculations.

But before that I want to thank everyone who helped to improve this tutorial: Specialist290 who proofread it and Kerolyov who showed a better 𝛥v map by metaphor and contributed to this chapter.

EVA to Transfer Crew

By the way, many of us wish we didn’t need to transfer crew with EVA when docked. But this kind of EVA was possible in the Apollo program. In case they couldn’t dock properly or open the docking tunnel, they’d have to go outside while connected to the lunar module’s life support system, enter the Command module and attach to its climate control (there were 2 sets of slots on space suits for safe re-connecting).


After bad docking, Apollo crew could still spacewalk from Lunar module (bottom) to Command module (top)

Stop Learning Math for Math Itself

Most people don’t like Math, and neither do I, despite showing you many formulae. I do all this math not out of passion for math itself, but for a purpose. This is my solution to learning anything: a compelling goal and putting in a reasonable effort to reach that goal. The reason why many think they don’t get math is simple: they don’t have a purpose. Thus, trying to learn by rote memorization is mostly a waste of time.

Here’s how you can easily understand the Tsiolkovsky equation.

Rocket Equation Is a Proportion

The equation is actually about ratios (fractions), not end values. This allows us to calculate fuel requirements and meters per second (𝛥v) from it for ships of any size:

m_0 = m_1 e^{\frac {\Delta v} {v_e}}

This is easier to see if we divide it by m1:

\frac {m_0} {m_1} = e^{\frac {\Delta v} {v_e}}

It is the ratio of start mass m0 to end mass m1. Quite simply, for every few meters per second increase in velocity you want to have, your ship needs to burn a proportion of its initial mass (propellant/fuel) from it’s rocket motor. How many “meters per second” the rocket velocity increases by for a given proportion of fuel burnt depends on engine efficiency (specific impulse or exhaust velocity). If you look at the fraction \frac {\Delta v} {v_e} it’s clear that when 𝛥v equals the exhaust velocity, (power is 1) then the ships initial mass is e times greater than its final mass. So to get a velocity increase (𝛥v) that is equal to your engine’s exhaust velocity the ship needs e times as much mass at the start as at the end of the burn.

There’s a even more intuitive way of thinking about it. What is the delta-v gained when the initial mass is twice the final mass i.e. when propellant mass burned is equal to half the ships initial mass? Take a logarithm of the previous equation:

\frac {\Delta v} {v_e} = ln \frac {m_0} {m_1} = ln2 \approx 0.693

So everytime a ship burns half its initial mass its increase in velocity will be equal to just less than 70% of its rocket exhaust velocity.

In the following diagram we have a Soyuz craft with a dry mass of 1 unit and 7 units of fuel (total of 8). If we burn 4 units of fuel we have halved its initial mass and so increase velocity by 0.69*v*:sub:e. We now have a total craft mass of 4 and if we burn half that (ie 2 fuel) we again increase velocity by 0.69*v*:sub:e for a total 𝛥v of 1.39*v*:sub:e. If we burn the remaining unit of fuel we again increase velocity by 0.69*v*:sub:e for a total 𝛥v of 2.08*v*:sub:e.


That’s all you need to remember. I’ll show how we get other forms of the equation.

Starting from our equation:-

m_0 = m_1 e^{\frac {\Delta v} {v_e}}

we can calculate the propellant mass burned for a given delta-v change by simply subtracting m1 from both sides:

m_0 - m_1 = m_1 e^{\frac {\Delta v} {v_e}} - m_1 = m_1 (e^{\frac {\Delta v} {v_e}} - 1)

Now, the method to estimate fuel requirements is simple:

  1. Divide maneuver 𝛥v by exhaust velocity (or specific impulse multiplied by g)
  2. Take that number for an exponent of e.

You get a number bigger than 1, which is a rate or percentage (like 1.5 is 150%), in which 1 is the mass of the ship itself (i.e. the dry mass), and the rest is the fuel.

Example: LV-909 has specific impulse of 390. You need to have 1000 m/s available for maneuvers.

e^{\frac {1000} {390 \times 9.81}} \approx 1.2987

So 29.87% of the mass the ship has in the beginning will be fuel used in the maneuvers.

Of course, we may have to add or remove stages with extra fuel tanks depending on the mission, but they have fractions of fuel mass versus structural and payload mass themselves, and the estimate is still good. For precise values, of course, we need to do multiple iterations of the 𝛥v equation, or, if good at maths, make an equation.

Lander Affordability Budget™

Now when we can estimate a maneuver cost, it’s easy to know if it’s worthwhile to build a separate lander.

On the one hand, a lander, which is lighter than the main ship, saves much fuel on maneuvers. Also, if we play by the same rules as real life space programs, we need to return the main ship and its crew home, which takes fuel, and hauling all this fuel back and forth is costly. On the other hand, the lander needs its own structure, decouplers, command pod, etc. If a maneuver is easy, it may be better to just land and return the whole ship.


The procedure is simple:

  1. Calculate how much fuel we need to add to the existing ship for it to undertake landing and take-off itself.
  2. Calculate the full mass of a separate lander capable of doing the landing and take-off.
  3. Compare the two figures.

To find the lander full mass you can just assemble one in VAB and use a mod such as Kerbal engineer to display its mass and delta-v.

I understand that when changing configuration we need to add or remove decouplers, docking ports, etc. But the overall mass does not change much.

Actually, we don’t need to calculate these numbers directly. Let’s solve a simple equation. Let’s say we need a ship with mass x to return to Kerbin from orbit around our target planet/moon. We now must calculate how much mass needs to be added to x to land and return to orbit before returning and then compare this to mass of a separate lander. We need to find a limit on lander full mass, Y from the 𝛥v we need.


The maneuver ratio is how much heavier the ship will need to be to perform the maneuver.

r = e^{\frac {\Delta v} {I_{sp} \times g}}

For landing the maneuver ratio tells us how much heavier the ship is when we add the fuel needed for the landing and return to orbit.

Ship plus lander full mass must be less than mass of the ship doing the landing itself.

x + Y < rx


Lander full mass (Y) has to be less than this:

Y < x (r - 1)

So if our returning ship mass is 7.168t (6.5t plus 668 kg of fuel), and we assume 1500 m/s of 𝛥v for landing and take-off, then the upper limit on our lander full mass is:

r = e^{\frac {1500} {390 \times 9.81}} \approx 1.480

Y < 7.168 \times (1.480 - 1) \approx 3.44

Lander full mass should be less than 3.44t. Our smallest lander mass was 2.87t, which is 570 kg smaller. As you can see, if we’d need to send the third kerbal to the surface, the third can would have added 600 kg, hence it would have been cheaper to land the whole ship.

You may ask, given such small weight gain, why did Apollo program use a rendezvous, then?

Compare to Real Life

If you look at Apollo program figures, the same maneuvers in real life cost much more. Trans-lunar injection costs 3000 m/s (instead of 850). Lunar orbit insertion requires 750 m/s (250 in KSP). Landing and take off cost 1600 m/s each (640 in KSP). Trans-Earth injection is about 750 m/s (250 in KSP).

Specific impulse of Apollo CSM was 314s. Its dry mass was 30t, hence trans-Earth injection required

30t \times (e^{\frac {750} {314 \times 9.81}} - 1) \approx 30t \times 0.276 \approx 8.27t

of fuel, and total mass was 38.27t. Maneuver ratio for lunar landing is

e^{\frac {3200} {314 \times 9.81}} \approx 2.826

Ship gets lighter 2.83 times at least, or fuel mass has to be 64.6% of total ship mass. Then the lander maximum mass is

Y = 38.27 (2.826 - 1) \approx 69.88

Lunar Module full mass was 14.6 tonnes, so it saved about 54 tonnes of mass. How much did this save for the whole mission? Well, we don’t count the structures, but estimating a low end is still interesting.

Low Earth orbit required 𝛥v of 9000 m/s. Trans-lunar injection - 3000 m/s, and lunar orbit - 700 m/s. 12,700 m/s in total. Specific impulse was, let’s say 400 on average (Saturn rocket did most of the work, and it had Isp = 410). So r = 25.444 (!), and multiplying this by 54 tonnes, we get 1373 tonnes. I guess, staging would have lowered this extra weight, as the proposed Nova rocket, for the single-ship mission, was “only” 800 tonnes heavier than Saturn V.

Just for comparison, current heavy-lifting rockets like Proton-M weigh 600-700 tonnes.


Rockets in their early namings: Comparison of Saturn IB (C-1) and Saturn V (C-5), 2800 t, with proposed Nova rocket (3600 t)


Proton-M rocket (700 t)

This is all for today. Next time I’ll show you a mission to Duna. Fly on budget and enjoy flying!

note:Images are either mine drawn for this tutorial, or made by NASA and in public domain.

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See also

One Giant Leap
a photo gallery of Apollo program (with comments in Russian)