Imagein a rocket with a central fuel/engine stack and four fuel/engine stacks of the same size and type attached with radial decouplers. Here are two scenarios:

Scenario 1 - The outer four fire together as your first stage. Then they are decoupled and the center one fires as your second stage.

Scenario 2 - The outer four and center one all fire together as your first stage. Fuel ducts are used to feed fuel into the center tank so that when the outer tanks are empty, you can decouple them and continue with just the center engine having full fuel. That is the second stage.

In both cases the second stage consists of one engine and a full tank of fuel. But the first stage is what I'm interested in. Everyone seems to think scenario 2 is the way to do it. But I was thinking about the math and I came up with this: scenario 2 gives you more thrust, but for less time. So you get 5/4 as much thrust, but since the four outer tanks are fueling all five engines your first stage lasts 4/5 as long. So how far does that get you? The distance should be acceleration times time squared. So 5/4 * 4/5 * 4/5, which seems to end up being only 4/5 as far as scenario 1.

Is that right or am I missing somthing? Do you actually get farther with the scenario 1 setup? I can't get to KSP right now to test this experimentally. I'd like to understand the math behind fuel ducting.

  • 1
    A couple things you're missing - first, atmospheric and gravity drag. When you're taking off from Kerbin you're losing a tremendous amount of energy just to break through the atmosphere, and although you're not thinking about it much, there's a lot of energy you lose to gravity drag when you're lifting off vertically. Getting out of the atmosphere as quickly as possible is usually considered more efficient (but there are limits to this as well). Commented Nov 1, 2013 at 13:46
  • @FooBarrigno Huh, I was told that trying to keep your vessel at approximately terminal velocity was the most fuel-efficient way to achieve orbit. Do you get more out of maximal thrust?
    – Sconibulus
    Commented Nov 1, 2013 at 13:52
  • @Sconibulus, nope, terminal velocity is the most fuel-efficient way. But for larger vehicles, it can take quite a bit of thrust to get to terminal velocity. Commented Nov 1, 2013 at 14:13
  • 1
    "Don't forget the books." You're not accounting for the mass of the fuel as you burn it in your acceleration and final velocity calculations. You can't do simple kinematics, but it's a relatively straight forward process to account for the fuel burned. Take a look at the rocket equation for more.
    – MBraedley
    Commented Nov 1, 2013 at 17:11
  • Actually, nevermind. In an ideal world, (i.e. not in a gravity well and without atmospheric drag), it doesn't matter how you fire, so the rocket equation doesn't really tell you anything. But it does make a difference.
    – MBraedley
    Commented Nov 1, 2013 at 17:36

3 Answers 3


Your question is actually about how fast you are going to spend the fuel of your first stage: Are you using four or five engines simultaneously to convert it into speed? When you use all 5 engines but throttle down to 80% thrust, it's exactly the same as if you would use 4 engines at 100% thrust. To answer this question, we first need to talk about what forces affect a rocket during lift-off.

There are two forces which prevent your rocket from getting into orbit.

  1. Gravity
  2. Atmospheric Drag

The first, gravity, is accelerating your rocket constantly downward until you have enough horizontal velocity to cancel it (achieved a stable orbit). The more time you spend affected by gravity, the more speed ("delta-v") does the gravity give you which you have to cancel by expending additional fuel. For that reason it is economically to accelerate early, so you get to your orbit faster and consume less acceleration from the gravity.

But there is also the second force: atmospheric drag. Atmospheric drag depends on atmospheric pressure and speed. The relationship with speed is quadratic. The faster you go, the more speed ("delta-V") you lose through air friction. That means it might not be so good after all to go too fast while you are still in the lower atmosphere.

So where do these two factors cancel out?

The ideal speed to balance atmospheric drag and gravity drag (assuming perfectly vertical ascent) is equal to the terminal velocity on the current atmospheric density. That speed depends on how aerodynamic your vehicle is.

When you go faster than this, you are wasting fuel on atmospheric drag. When you go slower than this, you are spending unnecessary fuel to fight gravity.

To get back to your initial question "what's better: serial staging or parallel crossfeed staging": It depends on your total thrust-to-weight ratio in the lower atmosphere. But my general experience is that a rocket gets higher with cross-feeding.

But what when you are already in orbit?

The truth is, it doesn't matter. The amount of delta-v you get per liter of fuel depends on the total weight of the ship and the average fuel-efficiency (Isp) of the engines you use. It doesn't matter how fast (through how many engines) you spend it. All that matters is to avoid having more mass than necessary (get rid of fuel tanks when they are empty). An orbital transfer stage with less engine power is more efficient, only because it tends to have a lower overall mass. This, however, is bought with longer burn-times to get the same amount of delta-v. Longer burn-times can sometimes mean less efficient burns because you can't hit your maneuver nodes that exactly, but this only matters in situations where a burn is very time-critical.

  • I disagree with your last two paragraphs about it not mattering when in orbit. It does. In orbit, if you crossfeed the center tank, it will get you further than if it wasn't crossfed. Now this may not make a big difference with small engines and tanks, but with large/jumbo tanks and large engines, it makes a big difference! Crossfeeding will get your more thrust by being able to use your center engine, and then being able to shed the mass of the other 4 while still having a full tank. I've learned this by creating asparagus stages for orbital transfer vehicles, it gets me much further.
    – Delorean
    Commented Dec 11, 2013 at 14:43
  • @XToro I think I addressed this with the sentence "All that matters is to avoid having more mass than necessary (get rid of fuel tanks when they are empty)". Cross-feeding will help you doing that. Btw: Your asparagus-staged orbital transfer vehicles would be even more efficient when you would omit all engines but the cental one. The others are just unnecessary weight.
    – Philipp
    Commented Dec 11, 2013 at 14:48
  • @Philipp Then that's a contradiction of "truth is, it doesn't matter", when in fact it does. And about my stages; not all my tanks have engines, only the 4th and 2nd last stages of the "orbit asparagus". They help push the mass of the extra fuel to hit my nodes better.
    – Delorean
    Commented Dec 11, 2013 at 15:21
  • @XToro the phrase "it doesn't matter" refers to the rephrased question "how fast you are going to spend the fuel of your first stage". And I also addressed the thing about accurately hitting maneuver nodes with the last sentence "Longer burn-times can sometimes mean less efficient burns because you can't hit your maneuver nodes that exactly, but this only matters in situations where a burn is very time-critical.". Interplanetary burns are rarely time-critical.
    – Philipp
    Commented Dec 11, 2013 at 15:37

So, I have been examining this issue both in KSP, and from the mathematics behind how rockets perform.

Simply put, in all cases, staging (specifically "asparagus" staging) is the best way to get the most Delta-V out of any rocket.

If you aren't familiar, check out the simple formula for The Rocket Equation

This equation describes how much Delta-V a given rocket has. So let's take a closer look...

Dv = Ve * Ln (Mt / Me)

Dv = The total Delta-V

Ve = propellant velocity

Ln = Natural Log function

Mt = total mass, Including propellant

Me = empty mass, without propellant

Starting with Mt / Me, it's important to note that although this part of the equation is measured in Mass, the units end up falling off and you end up with a simple ratio; the full to empty ratio. What this means is that it doesn't matter how big or small your rocket is, when calculating Delta-V, only how much of it is made up of Propellant.

The Ln function gives us a diminishing return for improved ratios.

Ve is the effective exhaust velocity of the propellant, often measured in meters per second, but most rocket engines list an Isp (specific impulse). Since Ve = Isp * G it's easy to plug this directly into the equation.

With either Isp or Ve, this does not describe the force put out by the engine in any way - just the efficiency of that engine - so a more powerful engine (given equal mass) does nothing to change your Delta-V. In reality a more powerful engine has more mass, which actually reduces the total Delta-V of the system.

Now... all this context is fine and all, but what about your question? Does using Staging actually improve Delta-V?

Yes, and here is why:

A single stage rocket uses the Rocket Equation: Dv = Ve * Ln (Mt / Me)

The same is true for any particular stage in a multi-stage rocket; that stage uses the same rocket equation:

Dv1 = Ve * Ln (Mt / Me)

Dv1 = Dv for stage 1.

Since any mission is the work of multiple stages, you can simply add the stages together to get the total Delta-V for the mission:

Given N Stages: Dvt = Dv1 + Dv2 + ... + Dvn

So - armed with the math, let's do some calculations!

Suppose you have a single stage rocket that is 80% propellant, with an effective exhaust velocity of 3234m/s (a Rockomax Mainsail in Vacuum). Remember, the actual mass doesn't matter - all we need is the ratio... But we'll plug in the masses to get there so our brains can keep up.

Dv = 3234m/s * Ln(5tonsFull / 1tonEmpty)

Dv = 3234m/s * Ln(5)

Dv = 3234m/s * ~1.61

Dv = ~5207m/s

Now, let's break this up into STAGES. Remember Dvt = Dv1 + Dv2 First Stage uses 50% (2 tons) of the available propellant:

Dv1 = 3234m/s * Ln(5tons / 3tons)

Dv1 = ~1652m/s

Second stage drops off 0.5 tons empty mass during decoupling:

Dv2 = 3234m/s * Ln(2.5tons / 0.5tons)

Dv2 = 3234m/s * Ln(5) (Doesn't this look familiar?)

Dv2 = ~5207m/s

Now total the stages:

Dvt = ~1652m/s + ~5207m/s

Dvt = ~6859m/s

Now, we can keep doing so for as many stages as we want (within the limits of the technology we have). Every time we divide the rocket into stages, we increase its delta-v. Asparagus staging is an expression of this mathematical principal.

  • Although the rocket in the video uses asparagus staging (at least it looks that way), it doesn't really explain why that's better than onion or parallel staging. It's more about the gravity turn than anything else.
    – MBraedley
    Commented Nov 4, 2013 at 12:58
  • It's not mean to be a replacement answer for Philipp's excelent explanation - so much as additional info, as Phillip mentioned the additional forces acting on a rocket during ascent. The first half of his post explains (in text) essentially the same thing the video explains (in video =p ). Commented Nov 4, 2013 at 19:47
  • Please don't post answers which rely on a linked video, because videos on YouTube often disappear after a while.
    – Philipp
    Commented Nov 16, 2013 at 17:24
  • @Philipp Answer improved... for more, check out my Youtube channel at youtube.com/users/korbitalmechanics Commented Nov 17, 2013 at 20:15
  • Your example leaves out a lot of details, which means it doesn't really reflect the problem space (namely serial vs. parallel staging). You're also effectively delivering less payload to orbit with the two stage rocket, or at least that's what it looks like.
    – MBraedley
    Commented Nov 20, 2013 at 17:24

"Less time" is what you want - "Less time to orbit" which necessitates "moving as fast as possible, as early as possible". Extra engine gives you the early boost.

That is to counter gravity losses.

Imagine this: a rocket like yours but with different engines. Two boosters have engines that are capable of lifting the entire stack and nothing else. Two next boosters have enough thrust to lift themselves and the center stage but not the first two. Then the center stage has enough thrust to lift just itself and nothing else, not accelerate, just hover over the launchpad.

Scenario 1. Fire the first pair of boosters. When they run out, dump them, fire the second pair, then when it runs out, fire the center stage. You didn't go anywhere, you kept hovering centimeters over the launchpad and you're now back down without fuel.

Scenario 2. Fire everything at once. First two boosters offset the gravity, then the second pair and the center provide acceleration. Then dump the two first, the side boosters offset gravity and thrust of the central provides acceleration. Dump the side ones and gravity losses are low enough you'll reach orbit on the central stage.

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