# How to find the optimum balance between amount of fuel and number of engines?

I'm coming from a background of Kerbal Space Program, so I'm not 100% sure all of this applies to real-life but I'm asking in the hopes someone might tell me that. Yes, I know, it's not real space travel and it doesn't accurately simulate all physics perfectly, but it's a good approximation and the core concepts are unchanged, so I consider it a good way to get introduced to concepts which one can then go and research further.

One thing that becomes pretty apparent from the start is that to get to orbit you need engines. To power these engines you need fuel, but fuel has mass, so you need more fuel and eventually things get pretty large and expensive. Each engine has a specific impulse and a craft (as we understand it to be a collection of payload, fuel, and engines) has a certain amount of delta-V, or the change in velocity possible.

Every maneuver, ascent, or powered landing the craft makes eats away at this budget of velocity change. From playing around it seems to be that making the following changes has the following effects:

• The thrust to weight ratio decreases. In orbit this doesn't appear to matter, but when landed it must be greater than 1 to have enough thrust to lift the craft.
• The amount of time the engines can burn for increases, but does this mean the delta-v only increases if you assume you are not ascending/landing and therefore do not need a TWR > 1?
• The TWR increases The thrust increases (and this usually increases the TWR, but not if the engine is exceptionally heavy in relation to the thrust produced), but the amount of time the engines can burn for decreases as more fuel is burnt in the same amount of time.
• I am unsure how delta-v changes here. It seems like adding more engines without adding more fuel does not increase it, it just means the same amount can be expended in a shorter time.

So when constructing a spacefaring vehicle (in reality), is there a mathematical or "proper" way to determine the optimum balance between fuel and engines? Furthermore, is such a method dependant on the celestial body/bodies you are within the Hill sphere of?

• Since you don't mention it...do you know about the rocket equation? en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation Feb 28, 2017 at 14:08
• Oh god, all that maths. I can't even say it's not rocket science. So I take it (I have no strong maths background) the rocket equation can be rearranged to allow me to find out the optimum balance of engines/fuel if I have a set of available existing engines with known properties? Feb 28, 2017 at 15:01
• @Leylandski: While you don't need to know all that math, you need to know $\Delta v =v_e ln { m_0 \over m_f }$ and $v_e = I_{sp} g_0$ - these two are all that binds high school level physics with rocket science; you can solve 90% of problems involving maths on this SE site with these and Newton's 3 laws + conservation of momentum.
– SF.
Feb 28, 2017 at 16:27
• Mandatory related XKCD Feb 28, 2017 at 21:28
• Related Feb 28, 2017 at 21:36

In general, you want the minimum number/thrust of engines that gets you off the launch pad safely -- in the real world this is typically a TWR of 1.2-1.4. The TWR increases rapidly as fuel is used, so if that minimum thrust level can be achieved it's almost always more effective to carry additional mass in fuel rather than engines.

You're correct that the thrust level doesn't affect delta-v at all -- the rocket equation (which is not very much math at all, really) doesn't have a thrust term. The only variable factors in it are the mass ratio $\frac {m0} {mf}$, which is maximized by carrying a lot of fuel, and the exhaust velocity $v_e$, which is a property of the engine type. In KSP, exhaust velocity is indicated via the specific impulse figure for an engine, which when multiplied by 9.81 gives the exhaust velocity in meters per second.

The other part of the rocket equation is the natural logarithm function $ln$ which reflects the diminishing returns of carrrying more fuel.

• More thrust reduces time spent at surborbital speeds, so less time for gravity losses. No idea how to calculate that though. Feb 28, 2017 at 15:35
• If the initial TWR is too great, the rocket reaches high speeds at low altitudes, and the savings from gravity loss are cancelled by drag increases (which go as the square of velocity). In addition, for real world rockets, there will be structural mass, thermal management, and crew comfort concerns. Feb 28, 2017 at 15:54
• The way around this is (usually) staging, not drop tanks -- you generally want to scale down your engines about the same time as you scale down your dry mass, for structural mass and crew/payload comfort reasons. Feb 28, 2017 at 16:00
• Usually you'll find that the smallest engine that gets you off the ground is the right one, if you're talking about a land-and-return mission. The mass of your ascent engine is dead weight that has to be carried by the descent stage, translunar stage, and launcher. One exception is if you're using a single stage for both descent and ascent; in that case you'll be overpowered for liftoff but you'll make up for it in simplicity and compactness -- this is also a place where drop tanks are worth considering. Feb 28, 2017 at 16:22
• @Leylandski in a vacuum what it basically comes down to is that you need to angle the thrust such that gravity is countered and all remaining thrust can go to building horizontal velocity. For example if you have a TWR of 1.41 the ship needs to pitch at 45 degrees, this results in 70.7% of the acceleration going to counteracting gravity and 70.7% building horizontal velocity (this is sine and cosine of the pitch). So a respectable fraction of the thrust can go to horizontal velocity even for low TWR. As horizontal velocity increases gravity is increasingly counteracted by centrifugal force. Mar 1, 2017 at 11:21