KSP Engine Efficiency Chart

[Seanner]

Spreadsheet answers below for those that don't care about the math, summary = use aerospikes, edit: upon further playing with values, it's more complicated...for regular interplanetary values at lowish TWR and dV, use whatever best lifts the required payload because the values are fairly close, but as dV increases the nuke gets wayyyyy better, but then as TWR increases the 48-7S gets wayyyyyy better....one doesn't need high TWR in space so the short answer is rockos for takeoff (high TWR) if you really want to, but aeros or LV-T30s + boosters are only very slightly worse because the boosters take your liquid usage out of the rocko's dV dominance range and are far simpler to build, plus 30s + boosters are cheap since solid fuel is cheap, so when in game money system is in place, a more "common sense" build is probably mathematically better too.

By setting the mission parameters TWR, gravity, and delta V, one can calculate which engine is best for the job.

I started with the mass (including fuel) that engines can lift at 2.0 TWR, and because of the fixed ratio of wet/dry mass for fuel tanks it was possible to solve the delta V equation entirely for payload:

p = L(9-F) / (8F) - E

where

L = total wet mass (liftoff)

F = e^(deltaV/(g*Isp)) = wet/dry ratio

It's possible to use this to solve for payload, and the most useful ratio in my opinion, at least until money comes into play, is fuel cost per payload.

Given deltaV = 4k, gravity = 9.81, and TWR = 2 for sea-level and 2k dV, 0.75 TWR for vacuum, we have:

Negative numbers imply that the engine cannot achieve the desired TWR.

I've recently learned that the atmosphere drops of super quick...a good solid booster phase will take you to near-vacuum Isps, so a more useful chart would've been a vacuum Isp at takeoff performance levels, but suffice to say 48-7Ss are superior, however marginally so: they don't really exceed other engines by much until the full 4.5k dV. Subtract 0.5-1.5k dV for solid boosters, and 0.5k because the interplanetary stage can take up the slack in achieving orbit (don't need 2.0 TWR to circularize...), and you have ~3k dV in liquid fuel, where at 2 TWR the aero has 2.77 fuel/payload and 48-7S has 2.35, and you need 4.6 of them to match the aero's payload. The LV-T30 is at 2.50 ratio and replaces 6.5 tiny rockos. Thus, you can safely build a "normal" rocket by achieving 2.0 TWR and 3k dV with fewest engines without worrying about min-maxing that last 5 drops of fuel in exchange for 300 more parts. Some concrete numbers at these parameters, starting with an interplanetary stage capable of landing on the mun and returning (0.75 TWR and 4k dV), given payload = 5 (big command pod and some other junk like science pod, goo pods, parachutes, batteries, etc.):

(note that mass is only additional mass of fuel and engines)

The LV-N costs too much, and the aero can't have stuff connect to it, so the LV-T30 wins the job of interplanetary and lander stage. If fuel cost is increased, the LV-N might be the best choice, or if the next stage's fuel cost increases too much because of increased mass of this stage. Given LV-T30 for the lander and its associated fuel, and decoupler now added as payload:

I'm sorting by fuel, presuming that is the true measure of liftoff efficiency...fuel is a valuable resource. But I don't want to use 57 engines, 9 engines, nor 12 engines, and in any case, the mainsail is cheapest. Since the payload exceeds 1 mainsail but isn't enough for 2, I had to use 8 BACCs to get to round trip Mun dV = ~8k. The mainsail only puts out 2350 at 1.91 if it burns through the BACC stage which would be slightly aerodynamically inefficient, so it will go a bit farther in dV but then lose a bit in underpoweredness...I suspect it's more or less enough to get to the moon though.

Another option if we consider cost in any way accurate is 3 skippers:

24-77s are cheap but the fuel cost is too high, so going back to the lower fuel options, 9 or 12 isn't too bad but T30s are way cheaper than aeros, so here's a setup with what was going to be 8-9 T30s, but in practice one needs control of one's ship so I went with 4 T45s and 6 radial T30s, each with 2 BACCs (!). This thing will get to the mun but it costs a bit...

Lastly, the optimal fuel and mass solution, 48-7S:

Edited November 10, 2013 by Seanner
Including sea level and vacuum

[numerobis]

Did you take into account the engine mass? That has to be subtracted from the payload.

[Seanner]

Yes, when I calculated the wet mass at launch, I used:

Mwet = [thrust / (TWR*g)] - Mengine

Which is a rearranging of TWR = thrust/weight, where weight is mg where m is everything except engine + engine.

edit: Actually I see the problem now that you mention it...I was originally calculating what it lifts using that formula, but now that's not actually the wet mass anymore, standby...

Edited November 10, 2013 by Seanner

[numerobis]

You might also want to look at the tables that tavert created. Basically the 48-7S wins almost everything.

[Seanner]

Having now fixed the error, I'm not sure tavert's tables are accurate then in the sense that what they are supposed to imply is really what they imply. Having fixed mine, It appears the aerospike still outperforms the 48-7S in fuel/payload. The mass of the engine itself isn't relevant. Basically, his table unfairly penalizes the aerospike for weighing more than the 48-7S. The point is that the aerospike lifts more payload with less fuel, regardless of how many aerospikes or 48-7Ss it requires to achieve such a lift. Engine performance as defined by a fuel/payload metric is independent of engine mass (even though it factors into its calculation), as it's not part of the payload, like you said! His table more closely shows reaction wheel performance and other things where the absolute total mass is important to know.

[numerobis]

Ah, OK, I think I see the distinction.

tavert is computing what engine you should use so that the total mass of the spacecraft at engine start is minimal. He computes the payload as being just the junk on top (science instruments, probe body, rovers, and that kind of silly stuff).

You're computing what engine you should use so that the propellent mass of the spacecraft at engine start is minimal. You define the payload as being the engine, plus the junk on top.

So you differ in two ways:

- propellent mass versus total mass

- payload includes the engine or not

[Dispatcher]

Your chart is limited to liquid fueled (and oxidized) rocket engines. It would be interesting to know how other engines compare: solid propellant, jet, mono propellant, ion.

[Seanner]

I think you wrote that backwards or you misunderstand: my calculation is actual payload (junk on top), and how much fuel that payload requires to lift it (fuel/payload), all to achieve say 4k dv at TWR=2. The mass of the engine itself is no longer significant. The way the chart works...pick a mainsail. It can lift 4.5 = big command pod and a 0.5 probe thing for example. Then attach 66 fuel to it (orange + half orange + random tanks to get close enough)...engineer redux should say very close to 4k @ 2.0. So it really does calculate fuel per payload, which to me is the true metric of engine efficiency (except in regards to reaction wheel performance or other silliness).

Here's the fixed equations if you want to verify, otherwise skip to the bottom for an emphasized summary:

Wet/Dry ratio, call it F (since it sort of represents fuel?) = e^[deltaV/(g*Isp)]

That's just a rearranging of deltaV = g*Isp*ln(wet/dry).

Then "mass at liftoff" (wet mass), call it L = thrust / (TWR*g)

Which is a rearranging of TWR = thrust/weight, splitting up weight into mass and gravity.

Then payload (P) = L - fuel wet (W) - engine mass (E)

Fuel dry is (D), so recall F = L/(total dry), and total dry is D + P + E

F = L/(total dry) = L / (D + P + E)

Now the problem is getting the dry fuel mass, but there's a trick in that the game sets D to W/9 (!) so:

D = W/9 = (L - P - E)/9

giving

F = L / [(L-P-E)/9 + P + E]

Looking at the denominator:

L/9 - P/9 - E/9 + P + E = L/9 + 8P/9 + 8E/9 = (1/9)(L + 8P + 8E)

so, together

F = 9L / (L + 8P + 8E)

FL + 8FP + 8FE = 9L

8FP = 9L - FL - 8FE

P = [L(9-F) - 8FE] / 8F = L(9-F)/8F - E, which is what I have in the sheet and OP though I have to edit the OP to reflect the subtraction of engine weight (the new sheet posted earlier is still correct!).

Now with payload, wet fuel is simply L - P - E (also was updated previously and so the image still stands).

Now, given the sciency-payload (NOT THE FUEL!!), and the wet fuel, we have the ultimate measure of efficiency -- how much fuel does an engine require to lift a payload for a given TWR and deltaV? That's what I'm showing. I believe my chart to be more useful, if less pretty... The mass of the engine is not important to its efficiency rating, though the text before tavert's suggests otherwise: "The question is for a given delta-V and payload, which engine gives the lowest total craft mass?". I don't care what the craft mass is! I care what the payload mass is and how much fuel it takes! (therefore I care about fuel per payload). If it takes a 70 teragram engine to lift it, so be it, provided it pulls that off with 1 gram of fuel (thus having extremely high Isp...suggesting that Isp is the be-all end-all of #s, except for that TINY constraint of TWR to actually reach orbit...this is why the nuke is a good choice in space but the aerospike is better on the ground).

If I'm in turn somehow misunderstanding tavert's data, my apologies...but at the moment I have to disagree that the tiny rocko beats the aero...

edit: And to respond to Dispatcher, the fuel of solid boosters is fixed, so those entries would be sort of meaningless. You get a set wet fuel and a variable TWR depending on what the deltaV goal is (and there is therefore a cap since deltaV is proportional to the log of wet/dry of solids (at an infinite number of them, otherwise you come up short from mass being wasted on payload, decreasing the overall wet/dry ratio)), which is especially bad because say you end up at TWR 1 then your deltaV is effectively 0-ish (it goes up as fuel burns off, but basically if you are in a hover you aren't really adding to your velocity which is supposed to be the interpretation of deltaV...you'll burn out the booster and be no closer to Mun). In practice, these high-powered (high TWR using their own weight, thereby bringing up the entire ship's TWR) boosters will fire WITH other engines, so you might find an efficient aerospike setup at 1.5 liquid TWR that nearly makes your dV, then just slap on some boosters to raise it to a take off of at least 2 or more TWR and some additional dV, and by the time the boosters burn off the aeros will have less fuel to lift bringing their own TWR up so it will all naturally kind of work out efficiently. I'm sure there's a more involved approach to solve for best liquid + solid usage but I don't have that at this time... Luckily it's kind of simple in practice => "need more dV and TWR" => booster++ (I think I've seen this in someone's signature..). Just going off the current listed costs, BACC's will outperform anything...so perhaps the algorithm is sort of reversed: given enough BACCs to get to TWR > 2, what did I get for dV and how many liquids do I need to provide the remaining dV beyond what the boosters are providing? Basically start with expendable engines and then figure out what kind of expensive "staying power" you need to reach orbit after they burn out. When I feel like it I might work on that problem too...

Edited November 10, 2013 by Seanner

[numerobis]

OK, now it's all clear and I agree with your direct conclusions.

I don't agree that you've computed the ultimate measure of efficiency . Rather, no measure is ultimate -- it all depends on what matters for the task at hand. You want to minimize the fuel burned (why?). Tavert wants to minimize spacecraft mass. Someone else might want to minimize part count, etc. Soon we'll start to care about minimizing the total cost.

If you have two stages, the second stage may want to burn more fuel if that lets it get lower total mass -- any extra mass on the second stage is more fuel required on the first stage. Neither your table nor tavert's figures directly optimize this, but you could use tavert's figures on the upper stage and yours on the first stage to get something reasonably close to optimal.

[Seanner]

You are absolutely correct about using his for the upper stage which I realized while modifying the OP. Saving mass IS important there because it saves "payload" relative to the lower stage. So I stand corrected. But I felt better claiming to have the ultimate formula, you see... Still, once the game has in-game costs, payload/cost or something like that could end up being the ultimate...