Oberth > Isp?

[FinalFan]

I just did a transfer burn from Kerbin to Jool with a vessel that has different LFO engines.  I used only the ones with better Isp but I was wondering if I would have done better to fire the other ones as well, counting on the increased Oberth effect to overwhelm the loss in fuel (burned less efficiently).  In this case I could have shortened a 3m27s burn to 2m35s. 

My instinct is that the amount you shorten the burn by doesn't matter a lot for this decision because the cost in fuel efficiency scales in proportion to how much thrust you are adding.  Is this instinct misguided? 
Is it always worth it when the engines are in the same general class?  (e.g. LFO generally gets 300-350 or 290-412 for the extreme cases) 
If so, is it always worth it, period?  (e.g. let's say hypothetically I have NERVS, LFO engines, and ore I can refine.  Should I make oxidizer?) 

[Guest]

I'm sure somebody could run the numbers, but I'm fairly certain that a one-minute longer burn will only cost single-digit m/s. You won't be that much higher at the end of the burn. The orbital period for LKO is about a half an hour; if your burn was 10 minutes as opposed to 1 minute then you'd see significant Oberth losses, and in that case you should split it into three 3.3 minute burns anyway.

If we're talking efficiency, I have to ask though, what are those other engines you might or might not burn for, and could you eliminate them from your design? Getting rid of that dry mass would net you a far bigger efficiency gain than a slightly shorter burn.

[Rayder]

It really depends on the use case. Since you cannot land on Jool you don't need to get close to it unless you want to build a low orbit station or something - or doing science (Jool's 'low/high space' boundary is 4,000 km).

Why it depends is because up to about 40,000 km (which lies just outside Vall's orbit, from memory), you can get huge fuel savings just by circularising up higher. The difference between a 40,000 km and 250 km circularisation is about 1,500 m/s deltaV. Going higher than 35--40 Mm doesn't save all that much fuel.

What's more interesting is that if you go much higher than 200,000 km it actually starts to take more fuel to circularise, so it's the sweet spot to do your insertion. Caveat: this window is HUGE! Going down to 100,000 costs <50dV more, and going up to 300,000 only costs <10dV more. Going all the way up to 2,000,000 km only costs ~200dV more.

This is the premise behind gate orbits. You're balancing the difference between your vessel's velocity and the velocity of your desired orbit. What's more, if you plan on returning to Kerbin then ~200,000 km is the most efficient orbit to leave from, so it could be a handy spot to setup your centre of operations. You'll be fighting Pol for the real estate though, so to avoid any accidents setup a little higher, maybe around 250,000 km. The difference is literally around ~2 dV for your insertion burn.

Ultimately I suppose you could calculate it, but I suspect your instinct is correct - there just isn't enough in it. Your best option I'd say is if you really wanted to maximise your savings using Oberth, consider using manoeuvre node splitter. You can split your insertion burn into two parts: 1st part just enough to capture into Jool's SoI and then 2nd to do the circularisation. I suspect you won't save all that much fuel, however.

Edited November 12, 2018 by Rayder

[FinalFan]

13 hours ago, Brikoleur said:

I'm sure somebody could run the numbers, but I'm fairly certain that a one-minute longer burn will only cost single-digit m/s. You won't be that much higher at the end of the burn. The orbital period for LKO is about a half an hour; if your burn was 10 minutes as opposed to 1 minute then you'd see significant Oberth losses, and in that case you should split it into three 3.3 minute burns anyway.

If we're talking efficiency, I have to ask though, what are those other engines you might or might not burn for, and could you eliminate them from your design? Getting rid of that dry mass would net you a far bigger efficiency gain than a slightly shorter burn.

Thanks everybody for all the answers!  It sounds like someone in low orbit (as opposed to a flyby) has considerable leeway on either side of periapsis in terms of taking advantage of the Oberth effect, so it probably helped to keep the less efficient engines turned off.  

Brikoleur, the less efficient engines were added for extra punch on the Tylo landing I wanted this vessel to be capable of.  It should have a local TWR of 1.45 fully loaded as I recall.  Tylo and Laythe are the only moons I'd need that kind of thrust for.  

That said, I am very silly and the thing is a gigantic beast.  Its name is "Admiral Perry" because of the muscular exploration it's meant to do.  It's designed to deliver up to two dozen kerbals in style to anywhere short of Eve within its 5,500 deltaV reach and carries enough ISRU to do it all over again without having to wait a year.  The more and less efficient engines in question are Rhinos and Vectors.  

@Rayder, I guess I was insufficiently clear when I said "transfer burn from Kerbin to Jool". What I meant was the actual burn that made me leave Kerbin and cross Jool's path.  But your information was very helpful.  I will certainly keep it in mind for my future maneuvers around Jool's orbit and other places.  

Edited November 12, 2018 by FinalFan
Rayder reply

[Vanamonde]

If you have mounted engines on the ship which you are not firing, you have much bigger fuel efficiency problems than Oberth math. Engines are heavy and engines that are not burning are dead weight which reduce the efficiency of the engines which are firing. That fact is going to overwhelm differences of engine efficiency in most applications. 

[FinalFan]

Just now, Vanamonde said:

If you have mounted engines on the ship which you are not firing, you have much bigger fuel efficiency problems than Oberth math. Engines are heavy and engines that are not burning are dead weight which reduce the efficiency of the engines which are firing. That fact is going to overwhelm differences of engine efficiency in most applications. 

Like I said in my reply to Brikoleur, I'll need it at the destination.  The extra engines are limited application but higher TWR compared to the main engines which are heavier but more efficient.  If we take as a given the single stage nature of the craft, is there still a trick I'm missing?

[Vanamonde]

6 minutes ago, FinalFan said:

Like I said in my reply to Brikoleur, I'll need it at the destination.  The extra engines are limited application but higher TWR compared to the main engines which are heavier but more efficient.  If we take as a given the single stage nature of the craft, is there still a trick I'm missing?

In most instances, if there are engines which can burn without incinerating what's below them, your net efficiency will be better if you burn them in addition to the fuel-efficient ones than if you just carry them as dead weight. Somebody who's better at the math could tell you where the cut-off point is at which this would become detrimental, but it's going to be an unusual scenario such as a very long, low thrust interplanetary transfer burn. 

[FinalFan]

57 minutes ago, Vanamonde said:

In most instances, if there are engines which can burn without incinerating what's below them, your net efficiency will be better if you burn them in addition to the fuel-efficient ones than if you just carry them as dead weight. Somebody who's better at the math could tell you where the cut-off point is at which this would become detrimental, but it's going to be an unusual scenario such as a very long, low thrust interplanetary transfer burn. 

Did a little math, and I reckon that, with the difference in Isp, the less efficient engines "wasted" enough fuel to run the efficient engines for about 1.875 seconds.  So the question becomes "is the difference in Oberth effect worth more than 1.875 seconds of free thrust?"  Based on my foggy recollection of the maneuver's total cost, and given Brikoleur's "single digit" estimate ... no, engine efficiency was better unless I completely screwed up my math or Brikoleur is off by a lot.  I calculated close to 20m/s so a minor mistake in the maneuver dV doesn't change the outcome.  

But there were a lot of moving parts in that calculation and it's not inconceivable that mistakes accumulated or my logic was flawed.  Or Brikoleur could be wrong.  Is there any relatively easy Oberth estimator around?  Either way, thanks for inspiring me to think about this.  

[edit:  got home and I think I messed up the fuel flow difference.  I think it's more like over 4 seconds of free thrust.] 

Edited November 13, 2018 by FinalFan
edit

[Rayder]

A craft using different ISP engines (for the same fuel type) effectively has the average ISP of all those engines.

Multiple engine specific impulse math

This thread discusses the deltaV of a craft with different ISP engines. You can effectively invert the equations to see what your remaining fuel mass will be if you use different types of engines. But overall your average ISP will be lower.

Granted, given your specific case things become more complicated since you're involving different engine types and also fuel synthesis. This is further complicated by how much fuel mass you have, how much extra LF you have for the nuclear engines, the mass of the craft etc. KSP makes things slightly easier since fuel and oxidiser have the same weight and density. So for a given fuel tank, you can work it out. Let's use the nuclear engine and the LV-909

Nuclear:
Mass - 3t
ISP - 800

LV-909:
Mass - 0.5t
ISP - 345

Since you're carrying both engines we will add them to the weight of our fuel tank. Let's use a Jumbo-64 tank.

Full Mass: 36t
Empty Mass: 4t

This gives us 32t of total liquid mass in the tank, 14.4t LF and 17.6t O. Now we can start working out our weights. Since the Nerv doesn't use the oxidiser, this means we have different 'wet' and 'dry' masses:

LV-909 Wet: 39.5t Dry: 7.5t
Nerv Wet: 39.5t Dry: 25.1t

So if we sub all of this data into our deltaV calculations, for our simple craft of 1 tank and two engines, each engine separately would give us the following results:

Nerv: 3,555 dV
LV-909: 5,617 dV

In this case, you can see that the LV-909 gets more delta V even though it's less efficient as it's burning away a lot more mass. BUT this is not what your situation is. You want to see if using more engines at once gets you more efficiency. For that, we now have to take in fuel usage rate into account. You also need to consider that as you're burning, some LF will be used by the Nerv so you'll have some excess oxidiser left over as dead weight. So let's re-sub all of this in one step, and calculate for both engines.

Averaged ISP: (800 + 345) / 2 = 572.5
Wet Mass: 39.5t
Dry Mass: 12.8t (5.3t of oxidiser remaining)

Now I REALLY hope I got all my math correct, because running both engines together results in a total deltaV for this craft at 6,322 m/s.

So yes, @Vanamonde is correct that using both could potentially net you better efficiency.

Edited November 13, 2018 by Rayder

[Snark]

2 hours ago, FinalFan said:

Thanks everybody for all the answers!  It sounds like someone in low orbit (as opposed to a flyby) has considerable leeway on either side of periapsis in terms of taking advantage of the Oberth effect, so it probably helped to keep the less efficient engines turned off.

Probably, yes-- in particular since the Isp gap between Rhino and Vector is fairly large.

Out of curiosity, did you use periapsis kicking?  That's one way to "have your cake and eat it too" when you have a low-TWR, high-Isp ship and want to take maximum advantage of Oberth effect.  Basic idea is, don't do the burn all in one piece.  Work out what your burn's going to be, then only burn for, say, a minute or so (i.e. a fraction of the total burn needed), which raises your Kerbin apoapsis... then coast around again and do a second burn at periapsis.  Depending on how long the needed burn is, you can make as many Kerbin orbits as you need, raising Ap each time, until you get your Ap raised up to 1000-2000 km or so (and therefore there's not much more dV left that you can add, without escaping entirely), so then you do one final burn that spends the entire remaining part of the burn.

If you've got a single-burn time of only a bit over 3 minutes, then your TWR's not all that low and you wouldn't need multiple passes in this case-- though you may be able to benefit from doing one periapsis kick.

(Another thing one can do-- if you're trying to pack in a lot of dV on your departing ship-- is to get to LKO and then do about 600-700 m/s of burn, thus putting it into a highly eccentric, low-periapsis elliptical orbit that's oriented correctly for the big burn you intend to make-- and then dispatch a fuel tanker to dock with it in that orbit and top off its tanks.  Gives you a "free" few hundred extra m/s at the destination.) 

[FinalFan]

On 11/12/2018 at 7:21 PM, Rayder said:

A craft using different ISP engines (for the same fuel type) effectively has the average ISP of all those engines.

Multiple engine specific impulse math

This thread discusses the deltaV of a craft with different ISP engines. You can effectively invert the equations to see what your remaining fuel mass will be if you use different types of engines. But overall your average ISP will be lower.

Granted, given your specific case things become more complicated since you're involving different engine types and also fuel synthesis. This is further complicated by how much fuel mass you have, how much extra LF you have for the nuclear engines, the mass of the craft etc. KSP makes things slightly easier since fuel and oxidiser have the same weight and density. So for a given fuel tank, you can work it out. Let's use the nuclear engine and the LV-909

Nuclear:
Mass - 3t
ISP - 800

LV-909:
Mass - 0.5t
ISP - 345

Since you're carrying both engines we will add them to the weight of our fuel tank. Let's use a Jumbo-64 tank.

Full Mass: 36t
Empty Mass: 4t

This gives us 32t of total liquid mass in the tank, 14.4t LF and 17.6t O. Now we can start working out our weights. Since the Nerv doesn't use the oxidiser, this means we have different 'wet' and 'dry' masses:

LV-909 Wet: 39.5t Dry: 7.5t
Nerv Wet: 39.5t Dry: 25.1t

So if we sub all of this data into our deltaV calculations, for our simple craft of 1 tank and two engines, each engine separately would give us the following results:

Nerv: 3,555 dV
LV-909: 5,617 dV

In this case, you can see that the LV-909 gets more delta V even though it's less efficient as it's burning away a lot more mass. BUT this is not what your situation is. You want to see if using more engines at once gets you more efficiency. For that, we now have to take in fuel usage rate into account. You also need to consider that as you're burning, some LF will be used by the Nerv so you'll have some excess oxidiser left over as dead weight. So let's re-sub all of this in one step, and calculate for both engines.

Averaged ISP: (800 + 345) / 2 = 572.5
Wet Mass: 39.5t
Dry Mass: 12.8t (5.3t of oxidiser remaining)

Now I REALLY hope I got all my math correct, because running both engines together results in a total deltaV for this craft at 6,322 m/s.

So yes, @Vanamonde is correct that using both could potentially net you better efficiency.

Everything you said seems basically true (I got slightly but not significantly different delta-V numbers when I checked you), but it's a very different situation than either my actual situation or my hypothetical NERV + other scenario.  The reason that burning both engines (or even the Terrier alone!) is better than NERV-only in your scenario is, as you said, because you eliminate a bunch of mass in the form of oxidizer.  But in the scenario, the question was whether ore should be refined exclusively into liquid for the NERV or partially into oxidizer for the LFO engine as well.  There should be no oxidizer left over since you simply wouldn't refine any that you weren't planning to use, and thus both should end at the same dry weight.  And in my actual scenario, of course, there was no issue of deadweight unusable fuel at all.  In both of these cases, clearly the efficient engine will always win out unless there is some other factor at play like a TWR requirement (such as on Tylo) or the Oberth effect as in the OP. 

I'm sure you were aware of the following, but it's worth noting that while the NERV and Terrier in your example make for a very simple average Isp (since they have the same thrust, you just add them up and divide by two), it can mislead readers into thinking it always works this simply.  In fact, you have to weight them by thrust.  For example, one Rhino (2000 thrust, 340 Isp) and one Vector (1000/315) average to ((2000 x 340)+(1000 x 315))/3000 = 331.67 Isp for the whole 3000-thrust combination. [edit:  @Zhetaan's post below gives the real, somewhat more involved, way to calculate average Isp.  The struck-through method will give you an approximation that I think should be fairly close as long as the engines' Isp isn't too radically different (i.e. Dawn or NERV vs. anything other than themselves).] 

Edited November 21, 2018 by FinalFan
Zhetaan

[Rayder]

When it comes to ore and converting to fuel, it's actually rather easy. Remember, when converting from ore to fuels the mass remains the same (with the exception of monopropellant).

Consider my simple craft scenario. If you change nothing else, it's more efficient to convert to just LiquidFuel and use the Nerv engine. Why? because LiquidFuel is 45% of the tank. Also, LiquidFuel and Oxidiser have the same mass, but are just used in different proportions.

How do we do it? Simple. Since the mass conversion from Ore to LF or Oxidiser is the same, you get the same out. Converting to LFO you'll get 14.4t of LF and 17.6t of O. But converting just to LF you'll get the whole 32t of LF. You get 2.2x the amount of LF as you'd get by doing an LFO conversion.

As a rough guess, you'd get ~7,700 m/s using the Nerv from the same amount of Ore, as it would take to fill the tank of LFO, where you'd get ~5,600 m/s using the LV-909.

I know my example craft isn't realistic, but consider that doing this for a useable craft in the game is much more complex, and it's dependent on the engines used, number of engines, tank sizes etc. There's simply too much to consider to give a generic answer.

Edited November 13, 2018 by Rayder

[FinalFan]

Well, now you've got me curious, @Rayder.  Two 2.5m ore tanks and one full Mk1 liquid tank makes the same 32 tons of fuel, but 4.25t dry weight instead of 4.  For fun, let's put an empty T400 on there to match the NERV's liquid tank, and with both engines that's 8t.  But there's also the converter—a 250, of course, to avoid wastage—but since we didn't calculate probe cores or anything for the original vessel I'll ignore cooling now.  So 12.25t dry and 44.25t wet.  Plug it in and...

Just over 10,000 delta-V. 

If we had ignored the different weight ratio of ore tanks and the need to add a converter, then it would have been extremely simple to figure out, and 7700 would be obviously way too low:  if the ONLY difference was 800 isp versus 345 isp, then we'd have 800/345 times the Terrier's ~5,600 m/s, over twice as much. 

1 hour ago, Snark said:

Probably, yes-- in particular since the Isp gap between Rhino and Vector is fairly large.

Out of curiosity, did you use periapsis kicking?  That's one way to "have your cake and eat it too" when you have a low-TWR, high-Isp ship and want to take maximum advantage of Oberth effect.  Basic idea is, don't do the burn all in one piece.  Work out what your burn's going to be, then only burn for, say, a minute or so (i.e. a fraction of the total burn needed), which raises your Kerbin apoapsis... then coast around again and do a second burn at periapsis.  Depending on how long the needed burn is, you can make as many Kerbin orbits as you need, raising Ap each time, until you get your Ap raised up to 1000-2000 km or so (and therefore there's not much more dV left that you can add, without escaping entirely), so then you do one final burn that spends the entire remaining part of the burn.

If you've got a single-burn time of only a bit over 3 minutes, then your TWR's not all that low and you wouldn't need multiple passes in this case-- though you may be able to benefit from doing one periapsis kick.

(Another thing one can do-- if you're trying to pack in a lot of dV on your departing ship-- is to get to LKO and then do about 600-700 m/s of burn, thus putting it into a highly eccentric, low-periapsis elliptical orbit that's oriented correctly for the big burn you intend to make-- and then dispatch a fuel tanker to dock with it in that orbit and top off its tanks.  Gives you a "free" few hundred extra m/s at the destination.) 

Periapsis kicking on a burn that's already less than two minutes from periapsis on each side?  Not my flavor of crazy.  It's come in handy, though, most notably when my first ion probe shipped out.  (Thankfully, all it has to do is hit a circular solar orbit for an asteroid detecting contract:  a relatively easy target to hit.) 

I like that fuel tanker idea, though.  It's certainly better than sending the tanker after a ship that's already outbound, which I did once. 

[Rayder]

Yes my scenario was still very simplistic, and ignored all the extra tanks, converters etc that you'd need to achieve it. But that's not needed to explain the concept.

Reality is, if you add all the extra weight and ore, yes your actual m/s will change - but the difference in efficiency will not. The Nerv would still be worse than the LV-909, and using them both would be better still. Even if you added an extra 100t to my example craft, you still get the same order:

Nerv: 854
LV-909: 881
Nerv + LV-909: 1,192

Nerv + Ore: 1,870

Remember that a core part of delta V calculating is the difference in the wet and dry weight. The only way to influence these numbers is to change the fuel capacities or the engines, which in my simple craft scenario we are not. By introducing fuel synthesis we are effectively modifying the available fuel which is why converting Ore the Nerv wins out.

Edited November 13, 2018 by Rayder

[FinalFan]

6 minutes ago, Rayder said:

Yes my scenario was still very simplistic, and ignored all the extra tanks, converters etc that you'd need to achieve it. But that's not needed to explain the concept.

Reality is, if you add all the extra weight and ore, yes your actual m/s will change - but the difference in efficiency will not. The Nerv would still be worse than the LV-909, and using them both would be better still. Even if you added an extra 100t to my example craft, you still get the same order:

Nerv: 854
LV-909: 881
Nerv + LV-909: 1192

Remember that a core part of delta V calculating is the difference in the wet and dry weight. The only way to influence these numbers is to change the fuel capacities or the engines, which in my simple craft scenario we are not. By introducing fuel synthesis we are effectively modifying the available fuel which is why converting Ore the Nerv wins out.

I understand the story you're telling.  The moral is, "if you've got oxidizer, burn it."  I agree—I think I did the math back when I first sent a spaceplane beyond LKO.  But I don't see how it's in any way relevant to the OP. 

[Rayder]

41 minutes ago, FinalFan said:

But I don't see how it's in any way relevant to the OP. 

As far as your original question goes, then no. There's no measurable efficiency loss by having a longer burn time. Even players who use Ion thrusters to go interplanetary, break up the burn into multiple stages do so for convenience, or to ensure the accuracy of the trajectory.

Best engine for interplanetary burns?

Edited November 13, 2018 by Rayder

[Tyko]

17 hours ago, Rayder said:

As far as your original question goes, then no. There's no measurable efficiency loss by having a longer burn time. Even players who use Ion thrusters to go interplanetary, break up the burn into multiple stages do so for convenience, or to ensure the accuracy of the trajectory.

I'm not sure this is correct. As @Brikoleur explains above, the longer your single burn is  the less efficient it is. Splitting it into multiple periapsis kicks is more efficient.

[EastWindCC]

On 11/13/2018 at 9:21 AM, Rayder said:

A craft using different ISP engines (for the same fuel type) effectively has the average ISP of all those engines.

Multiple engine specific impulse math

This thread discusses the deltaV of a craft with different ISP engines. You can effectively invert the equations to see what your remaining fuel mass will be if you use different types of engines. But overall your average ISP will be lower.

Granted, given your specific case things become more complicated since you're involving different engine types and also fuel synthesis. This is further complicated by how much fuel mass you have, how much extra LF you have for the nuclear engines, the mass of the craft etc. KSP makes things slightly easier since fuel and oxidiser have the same weight and density. So for a given fuel tank, you can work it out. Let's use the nuclear engine and the LV-909

Nuclear:
Mass - 3t
ISP - 800

LV-909:
Mass - 0.5t
ISP - 345

Since you're carrying both engines we will add them to the weight of our fuel tank. Let's use a Jumbo-64 tank.

Full Mass: 36t
Empty Mass: 4t

This gives us 32t of total liquid mass in the tank, 14.4t LF and 17.6t O. Now we can start working out our weights. Since the Nerv doesn't use the oxidiser, this means we have different 'wet' and 'dry' masses:

LV-909 Wet: 39.5t Dry: 7.5t
Nerv Wet: 39.5t Dry: 25.1t

So if we sub all of this data into our deltaV calculations, for our simple craft of 1 tank and two engines, each engine separately would give us the following results:

Nerv: 3,555 dV
LV-909: 5,617 dV

In this case, you can see that the LV-909 gets more delta V even though it's less efficient as it's burning away a lot more mass. BUT this is not what your situation is. You want to see if using more engines at once gets you more efficiency. For that, we now have to take in fuel usage rate into account. You also need to consider that as you're burning, some LF will be used by the Nerv so you'll have some excess oxidiser left over as dead weight. So let's re-sub all of this in one step, and calculate for both engines.

Averaged ISP: (800 + 345) / 2 = 572.5
Wet Mass: 39.5t
Dry Mass: 12.8t (5.3t of oxidiser remaining)

Now I REALLY hope I got all my math correct, because running both engines together results in a total deltaV for this craft at 6,322 m/s.

So yes, @Vanamonde is correct that using both could potentially net you better efficiency.

Obviously,there's something wrong.The total △V is about 4236 m/s.

I made a excel table and calculated everything related to Liquid Fuel/Oxidizer consumption.It looks like this.

	LV-N	LV-909	engines	3.5t
thrust	60kN	60kN	tank	4t
isp	800s	345s	liquid fuel	14.4t
LF	0.0076531t/s	0.0079858t/s	oxidizer	17.6t
OX	0	0.0097604t/s	wet mass	39.5t
mass	3.0t	0.5t	remaining ox	8.612766t
			dry mass	16.112766t
average isp	482.1s		wet/dry	2.4514723
			△V	4236m/s

I got the average isp in this way:   

isp=dI/dm

dI=F*dt and dm=m*dt,   where m is propellant consumption per second.

so isp=F/m

then convert m/s to s

You can't simply get (800+345)/2.

What's more,my calculations shows the higher thrust LV-N is,the lower ΔV you'll get.

Edited November 17, 2018 by EastWindCC
typo

[Rayder]

I don't know what to tell you. I checked in Kerbal Engineer and my test craft gives 4,239 dV and an Isp of 482. My math was wrong, and I checked the forum post I linked. You're right, the Isp is not simply (a+b)/2. But the math is complex, and I guess it's up to you whether it's worth the effort. Me personally, I just use Kerbal Engineer. Saves me the headache haha.

[Zhetaan]

9 hours ago, Rayder said:

You're right, the Isp is not simply (a+b)/2. But the math is complex [...]

I wouldn't go that far.  I agree that the maths are a bit more challenging than it would seem at first glance, but they're by no means excessively difficult.  The main thing is to remember that engines are not measured only by exhaust efficiency, but also by exhaust quantity (which ties to mass flow rate).  In order to combine the performance of two different engines, you need to know how efficient they are according to the fraction of propellant they contribute to the total thrust.

For the Nerv and the Terrier, even though they have the same thrust, the Nerv uses a lot less propellant to achieve that thrust (because it is more efficient).  That means that for a combination of the two, it is actually contributing less than half (quite a bit less, actually) to the fuel flow.  But that should make sense when you think about it:  if the Terrier is less efficient, then you ought to expect it to use up a lot more propellant to achieve the same effect as the Nerv.  But if it is contributing more to the fuel flow, then doesn't its efficiency carry greater weight?  If that's the case, then we should expect the combined efficiency to be somewhat closer to the Terrier value than to the Nerv value.

For actual numbers, I'll use the wiki values so you can see where @EastWindCC's number comes from (welcome to the forum and thanks for contributing!):

Nerv:  800 s I_sp and 60 kN vacuum thrust
Terrier:  345 s I_sp and 60 kN vacuum thrust

Thrust equals the mass flow rate times the effective exhaust velocity, which itself is equal to I_sp * g₀ (9.80665).

Nerv mass flow rate:  60000 N / (800 s * 9.80665 m/s²) = 7.648 kg/s
Terrier mass flow rate:  60000 N / (345 s * 9.80665 m/s²) = 17.734 kg/s

If these numbers don't look familiar, remember that LFO (and LF and O separately) has a density of 5 kg/L and the readout is in litres, so divide by five to get the resource panel value (total, that is--it's split between the LF and the O on the panel).

The total fuel flow is the sum, or 25.382 kg/s.  The Nerv supplies 800 s * (7.648 / 25.382) (approximately 30%) of the specific impulse, for an effective contribution of 241.053 s of the combined I_sp, and the Terrier supplies 345 s * (17.734 / 25.382) (approximately 70%) of the specific impulse, for an effective contribution of 241.046 s of the combined I_sp.  The average I_sp is the sum of the contributions, or 482.1 s.

One oddity should be readily obvious:  each engine contributes half to the average specific impulse in spite of the 70/30 split to the fuel flow.  This is something that arises because the Nerv and the Terrier happen to produce the same vacuum thrust:  if we had used engines with different thrusts, the numbers would have worked out differently.  However, this illustrates the important difference between the two:  as stated earlier, the Nerv develops the same thrust as the Terrier by use of less than half the propellant, which is the practical effect of its having more than double the efficiency.

Another oddity is less obvious:  482.1 s is a mean value between 800 and 345, but it's not the arithmetic mean as one would normally calculate by adding and then dividing the sum by two.  It's the harmonic mean, which is a different kind of average but still mathematically relevant.  If you know how to use it, then you can compare any two engines regardless of thrust and without needing to figure the fuel flow.  For example, let's say you wanted to run a Terrier with a Reliant for some absurd reason.  The Reliant develops 240 kN of thrust at 310 seconds, and the Terrier, as we've seen, develops 60 kN of thrust at 345 seconds.  Divide the I_sp by the thrust to get the unlikely unit of seconds per kilonewton, which for the Reliant is 1.2917 and for the Terrier is 5.75.  Take the harmonic mean, which is 2.1095, and multiply by the average thrust (150 kN) to get 316 seconds, which is both the mean specific impulse and an excellent justification for why first-stage engines belong in the first stage and vacuum-stage engines belong in space.

Edited November 20, 2018 by Zhetaan

[FinalFan]

10 hours ago, Zhetaan said:

I wouldn't go that far.  I agree that the maths are a bit more challenging than it would seem at first glance, but they're by no means excessively difficult.  The main thing is to remember that engines are not measured only by exhaust efficiency, but also by exhaust quantity (which ties to mass flow rate).  In order to combine the performance of two different engines, you need to know how efficient they are according to the fraction of propellant they contribute to the total thrust.

For the Nerv and the Terrier, even though they have the same thrust, the Nerv uses a lot less propellant to achieve that thrust (because it is more efficient).  That means that for a combination of the two, it is actually contributing less than half (quite a bit less, actually) to the fuel flow.  But that should make sense when you think about it:  if the Terrier is less efficient, then you ought to expect it to use up a lot more propellant to achieve the same effect as the Nerv.  But if it is contributing more to the fuel flow, then doesn't its efficiency carry greater weight?  If that's the case, then we should expect the combined efficiency to be somewhat closer to the Terrier value than to the Nerv value.

For actual numbers, I'll use the wiki values so you can see where @EastWindCC's number comes from (welcome to the forum and thanks for contributing!):

Nerv:  800 s I_sp and 60 kN vacuum thrust
Terrier:  345 s I_sp and 60 kN vacuum thrust

Thrust equals the mass flow rate times the effective exhaust velocity, which itself is equal to I_sp * g₀ (9.80665).

Nerv mass flow rate:  60000 N / (800 s * 9.80665 m/s²) = 7.648 kg/s
Terrier mass flow rate:  60000 N / (345 s * 9.80665 m/s²) = 17.734 kg/s

If these numbers don't look familiar, remember that LFO (and LF and O separately) has a density of 5 kg/L and the readout is in litres, so divide by five to get the resource panel value (total, that is--it's split between the LF and the O on the panel).

The total fuel flow is the sum, or 25.382 kg/s.  The Nerv supplies 800 s * (7.648 / 25.382) (approximately 30%) of the specific impulse, for an effective contribution of 241.053 s of the combined I_sp, and the Terrier supplies 345 s * (17.734 / 25.382) (approximately 70%) of the specific impulse, for an effective contribution of 241.046 s of the combined I_sp.  The average I_sp is the sum of the contributions, or 482.1 s.

One oddity should be readily obvious:  each engine contributes half to the average specific impulse in spite of the 70/30 split to the fuel flow.  This is something that arises because the Nerv and the Terrier happen to produce the same vacuum thrust:  if we had used engines with different thrusts, the numbers would have worked out differently.  However, this illustrates the important difference between the two:  as stated earlier, the Nerv develops the same thrust as the Terrier by use of less than half the propellant, which is the practical effect of its having more than double the efficiency.

Another oddity is less obvious:  482.1 s is a mean value between 800 and 345, but it's not the arithmetic mean as one would normally calculate by adding and then dividing the sum by two.  It's the harmonic mean, which is a different kind of average but still mathematically relevant.  If you know how to use it, then you can compare any two engines regardless of thrust and without needing to figure the fuel flow.  For example, let's say you wanted to run a Terrier with a Reliant for some absurd reason.  The Reliant develops 240 kN of thrust at 310 seconds, and the Terrier, as we've seen, develops 60 kN of thrust at 345 seconds.  Divide the I_sp by the thrust to get the unlikely unit of seconds per kilonewton, which for the Reliant is 1.2917 and for the Terrier is 5.75.  Take the harmonic mean, which is 2.1095, and multiply by the average thrust (150 kN) to get 316 seconds, which is both the mean specific impulse and an excellent justification for why first-stage engines belong in the first stage and vacuum-stage engines belong in space.

Wow!  I'll have to remember that somehow.  But is it fair to say that the error (produced by doing it the wrong way) increases dramatically, perhaps exponentially, as the Isp difference increases?  I compared the same engines (Terrier + Reliant) using the method I had formerly thought was accurate and came up with 317 (vs. 316, tiny difference).  Comparing a Mastodon + Wolfhound setup seems to yield about 310 versus 316.5 the wrong way.  (6.5 is still relatively tiny compared to 90.4 difference between NERV/Terrier calculations despite the Isp difference being over 25% as large)

Edited November 21, 2018 by FinalFan
replaced "increases radically" in case that could be mistaken for a math term

[Zhetaan]

5 hours ago, FinalFan said:

But is it fair to say that the error (produced by doing it the wrong way) increases dramatically, perhaps exponentially, as the Isp difference increases?

Actually, no.  I compared a Rhino with a Nerv using your method of weighted averages and got 353.398 seconds of combined specific impulse.  I used my method and got 345.791 seconds.  I also tried the Rhino/Vector combination you gave as an example:  your value was 331.67 seconds and mine is 331.237 seconds.  A Mammoth and a Dawn gave 316.942 seconds your way, and 315.146 seconds my way.  Fundamentally speaking, your method will always be high, and that is because you are using an arithmetic mean.  The arithmetic mean is always higher than the harmonic mean for reasons I will explain below, but the reason that you get so close anyway is because of the weight you give to your average.  The general form of your method is this:

(T₁I₁ + T₂I₂) / (T₁ +T₂)

Where T₁ is the first engine's thrust, I₁ is its I_sp, and T₂ and I₂ are the second engine's thrust and I_sp, respectively.  After a bit of algebra, my general form is this:

(T₁I₁I₂ + T₂I₂I₁) / (T₁I₂ + T₂I₁)

If you look closely, my method is (at least by this arrangement) very similar to yours in that it differs only by what I will call a pseudo-term of I₂ + I₁ in both the numerator and the denominator.  I call it a pseudo-term because it cannot be isolated from my form to get yours, which is why they are unequal.  But ignoring unwanted terms when one cannot otherwise be rid of them is also a time-honoured and long-practised method of approximation in higher mathematics, and this is why your method is usually a very good estimate when it compares very different engines.

However, this actually makes the approximation worse the closer two engines are to one another in thrust.  The reason it breaks down with the Nerv/Terrier combination is because that specific pair has identical thrust:  in your method, that allows the T values to divide out of the formula to leave a simple (I₁ + I₂) / 2 average--the weights are, in a sense, cut loose, and your formula becomes the thing that you were trying to avoid in the first place.  In mine, the T values also divide out ... but I don't weight it with thrust.  I normalise it with thrust and weight it with I_sp.  Of course, that means that my formula simplifies when the I_sp values are equal, but if they are, it simplifies to only I_(sp) regardless of the thrusts, which is one way of saying that I won't need a formula to tell me what the average I_sp is between two engines with the same I_sp.

As promised, here is an explanation for why the arithmetic mean will always be higher than the harmonic mean.  There are a few ways to calculate the harmonic mean that are all mathematically equivalent.  For example, you can take it as the reciprocal of the arithmetic mean of the reciprocals of the numbers you're averaging (that sounds complicated but it's the easiest way if you have a '1/x' key on your calculator).  For ease of understanding, suffice it to say this is probably the most straightforward calculation:  the harmonic mean of two values is equal to their product divided by their arithmetic mean.  This implies that the harmonic mean of two numbers a and b equals 2ab / (a + b).  If we start by assuming that the difference between a and b is zero (that is, they are equal), then we get 2a² / 2a = a, which we would expect.  All averages lie between the values they average, so if the values are equal, then the average must be equal to them.  On the other hand, if we hold a constant and allow b to increase arbitrarily, an interesting pattern develops:  the greater b is compared to a, the more it dominates the sum in the denominator.  Eventually, the difference is so great that a + b is approximately equal to b, which is seen, for example, when a = 1 and b = 1000000:  1000001 is approximately equal to 1000000.  If we assume that relationship to hold for ever higher values going to infinity, then we can substitute b for a + b ... which does two things.  First, it follows the time-honoured tradition of approximation, and second, it causes the b values to divide out and leave the harmonic mean equal to 2a--I won't bore you with the reason why you can divide infinity by infinity, but there's a way to do it using calculus.  As an example of this, using a = 1 and b = 1000000, the harmonic mean is about 1.999998, or two millionths short of 2 and thus approximately equal to 2a.

By contrast, the arithmetic mean works in the opposite way:  with two values a and b, the formula is (a + b) / 2.  If we take a and b to be equal, then the equation becomes 2b / 2 = b (or a if you prefer, but let me make my point), which, again, we expect.  If we hold a constant and let b increase arbitrarily as before, then we can substitute b for a + b as before, which gives the formula b / 2.  Unlike the harmonic mean, where if we hold a constant, then 2a is just as constant, in the arithmetic mean, the value of the mean is never less than half of b and so it must increase as b increases.

These extremes tell us something about how the formulae work, even though they are not practical themselves:  in these examples, the harmonic mean has a ceiling above which it will not go, and the arithmetic mean has a floor below which it will not go, and this is why the harmonic mean is always less than the arithmetic mean.

Edited November 21, 2018 by Zhetaan

[FinalFan]

@Zhetaan Wow!  Thank you for the extremely educational post.  I had been secretly wondering what the harmonic mean was.  Referring to the original question, you did answer it.  I believe I had been misled by reading too much into my three example data points that had multiple variables. 

Clearly the biggest difference between my oversimplified model and the correct one is illustrated in extreme cases that expose the difference in limits that are approached, 2a in one case and b/2 in the other.   

The most non-intuitive effect (for me) of doing things the correct way comes in if the high thrust and high Isp engines are the same, because I don't instinctively consider fuel flow and as conditions get more extreme the fuel flow on low Isp (higher fuel flow) engines will tend to overwhelm the high Isp (lower fuel flow) engine even if it's the lower thrust one.  But I suppose this is what causes the upper limit on combined Isp in the first place.  

I read your post at work, thought about it from time to time, and I'm back home now.  I can't remember the last time I rewrote a post so many times based on increased understanding, even if it's partly due to divided attention.