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Staging Optimally: wanna help me with math?

Asked by
Cunjo Carl,

- 2

Staging Optimally: wanna help me with math?

Asked by
Cunjo Carl,

Posted (edited)

Math is part of the fun!

Edit: It's possible I've answered my own question (see the reply) but a second set of eyes would still be super appreciated, if it sound like a fun puzzle to solve through.

I feel so close, and yet so far... I was honestly expecting to figure out the problem while rubber-ducking this post together, but alas no such luck. Best of luck on your analysis, and thanks in advance for the help or consolations! Oh, and as a personal request, thanks for poking holes in my equations, but please don't poke a hole you're not willing to fill back in

0. PremiseSo, we have a lifter that'll get m

_{craft}tons to LKO, and a final unpowered reentry stage weighing m_{pod}tons to finish the trip, and we want to maximize the amount of fun we can have in between.By fun, I mean ΔV with a given TWR

_{min}!We should be able use the rocket equation to find the optimum size of each stage relative to the next (so long as we're using the same engine specs everywhere.) Let's solve for the deltaV of a specific case, and generalize from there.

Yeah, that's the craft we're describing!

1. ΔV of a single stage, given masses and TWRs:We'll start with the rocket equation, and plug in values and stipulations specific to KSP to make it more convenient for our case.

ΔV

_{stage}= -g_{0 * }I_{sp * }ln(m_{empty}/m_{full}) wherem

_{empty}=^{1}/_{9 * }m_{tanks}+ m_{engines}+ m_{payload }-> (next line, substitute an equation for the fueltanks: m_{tanks})m

_{tanks}= m_{fullcraft}- m_{engines}- m_{payload }(everything that's not engines or payload is fuel) -> (substitute mengines, mpayload)m

_{empty}=^{1}/_{9 * }m_{full}+^{8}/_{9 }_{* }m_{engines}+^{8}/_{9 * }m_{payload}mengines = m

_{full}_{* }TWR_{min}/TWR_{eng}(Our engines must be this large to generate our min TWR)mpayload = m

_{full}_{* }(m_{payload}/m_{full}) .... please take my word! -> (next line, substitute mengines, mpayload. Factor terms)m

_{empty}=^{1}/_{9 * }m_{full }+^{8}/_{9 }_{* }m_{full * }TWR_{min}/TWR_{eng}+^{8}/_{9 * }m_{full *( }m_{payload}/m_{full}) -> (next line, factor out m_{full})m

_{empty}= [^{1}/_{9}+^{8}/_{9 * }TWR_{min}/TWR_{eng}+^{8}/_{9}_{* }(m_{payload}/m_{full}) ]_{* }m_{full}-> (next line, finally plug back in to deltaV equation from up top)ΔV

_{stage}= -g_{0 * }I_{sp * }ln( [^{1}/_{9}+^{8}/_{9}_{* }TWR_{min}/TWR_{eng}+^{8}/_{9}_{* }(m_{payload}/m_{full}) ]_{* }m_{full}/ m_{full}) -> (next line, cancel the m_{full})ΔV

_{stage}= -g_{0 * }I_{sp * }ln(^{1}/_{9}+^{8}/_{9}*TWR_{min}/TWR_{eng}+^{8}/_{9 * }(m_{payload}/m_{full}) ) (Hooray!)Great! We're not treading any new ground yet, but we can describe our deltaV in terms of known convenient variables. Let's march on.

Also, decouplers and other goodies can be accounted for alongside the engines in a similar way, but we'll neglect them for now.

Y axis: DeltaV for this stage in m/s. X axis: mass ratio of this stage's payload to its full mass (unitless).

We see 2000 m/s deltaV for a TWR=2 terrier stage that's 2.75 times bigger than its payload -> Sounds reasonable! This number will later turn out later to be our optimal stage ratio given our TWRs, and I may have fudged the TWRs a bit to make it to work out to about e just for the heck of it.

2. Extend ΔVfrom individual stages to a full craftWe'll bring in new equations for how multiple stages scale in deltaV and mass and use them to calculate deltaV for an entire craft worth of stages.

Posit: Stages will be geometrically scaled when optimally sized, and thus with uniform ΔV

_{stage}(mNum

_{payload}/m_{full})^_{stages}= (m_{pod}/m_{craft}) (In other words, the mass ratio of each stage multiplied together enough times will equal the mass ratio of our full craft)ΔV

_{craft}= Num_{stages * }ΔV_{stage}(Eq. 2) Yep, namin' this one. Just this oneReorganizing the top equation gives(1/Num

(m

_{payload}/m_{full}) = (m_{pod}/m_{craft})^_{stages}) -> (next line, plugging this into the ΔV_{stage }equation and that into Eq. 2)ΔV(1/Num

_{craft}= - g_{0 * }I_{sp * }Num_{stages * }ln(^{1}/_{9}+^{8}/_{9}*TWR_{min}/TWR_{eng}+^{8}/_{9 * }(m_{pod}/m_{craft})^_{stages}) )Neat, now we can extend our equation to a full craft. By plotting ΔV

_{craft }while varying Num_{stages}we can visually determine the optimal number of stages!Y axis: DeltaV for craft in m/s -> ΔV

_{craft ... }X axis, the number of stages in this craft -> Num_{stages.}The craft was declared to have a final pod of mass 1ton, and a starting (LKO) mass of 100 tons.

The numbers seem mostly believable, too. We can also double check our prediction from the original "posit" assumption is self consistent. At the optimal Num

_{stages }, (m_{payload}/m_{full}) (which is the stage mass ratio) is always the same irrespective of (m_{pod}/m_{craft}) (The craft mass ratio) ... Success! Just as an aside, this result seems weird given how the craft mass ratio is stuck in a logarithm with a summed term, but it's something we should expect physically and the equation somehow bears it to be true.This graph also consistent with our previous graph, where the dots I've highlighted on each graph correspond to the same point, and reassuringly 8998 ~ 4.58*1975 (or the deltaV of our craft is the number of stages times the deltaV per stage) (1/100)^(1/.458) = ~.364, the value of evaluation for the previous graph.

3. Generalize to a form that's irrespective of craft size ... Results seem mostly good.We'll take our equations for the specific craft and generalize them to work for any craft of any size. When optimally staging, the deltaV of a craft should scale linearly with the e-fold scale of its mass ratio, so we'll make an equation for this. The value can be viewed as a mass-to-deltaV efficiency factor. In another field, we'd probably call it our 'goodness factor'.

(mNumln(m(1/Num

_{payload}/m_{full})^_{stages}= (m_{pod}/m_{craft}) (Starting with the first equation from part 2. Next line, flipping and taking a log (ln for convenience))ln(m

_{pod}/m_{craft}) = Num_{stages *}_{payload}/m_{full}) (Next, dividing this from the final equation from part 2.Also, substituting the stage mass ratio (m

_{pod}/m_{craft}) back in to the logarithm in the place of (m_{pod}/m_{craft})^_{stages}))ΔV

=ln(m

_{craft}- g_{0 * }I_{sp * }Num_{stages * }ln(^{1}/_{9}+^{8}/_{9}*TWR_{min}/TWR_{eng}+^{8}/_{9 * }(m_{payload}/m_{full}) ) (Next line, canceling Num_{stages})ln(m

_{pod}/m_{craft}) Num_{stages *}_{payload}/m_{full})ΔV

=

_{craft}- g_{0 * }I_{sp }_{* }ln(^{1}/_{9}+^{8}/_{9}*TWR_{min}/TWR_{eng}+^{8}/_{9 * }(m_{payload}/m_{full}) ) (Finally inversing the pod/craft in the log on the left to orient it correctly with deltaV)ln(m

_{pod}/m_{craft}) ln(m_{payload}/m_{full})ΔV

=

_{craft}g_{0 * }I_{sp }_{* }ln(^{1}/_{9}+^{8}/_{9}*TWR_{min}/TWR_{eng}+^{8}/_{9 * }(m_{payload}/m_{full}) )ln(m

_{craft}/m_{pod}) ln(m_{payload}/m_{full})And that's it! The left is the deltaV of your craft relative to its scale (in e-fold), which should be constant and irrespective of the craft's scale, and we see to be the case on the right. The right only has properties that pertain to individual stages, nothing about the craft. Plotting the left in terms of stage relative mass (m

_{payload}/m_{full}) creates a graph which seems pretty reasonable, and I've checked parts of it to be true to the best of my ability to tell using KER. Its roots are in the right theoretical places, and the maximum moves in physically sensible ways with different TWRs. I'm definitely not convinced by the values near zero, though. They seem much too high to me. Could still be fine.Y axis: deltaV in m/s for every e-fold size of your craft -> ΔV

_{craft}/ ln(m_{pod}/m_{craft}) .... X axis: relative size of each stage (with its payloads) relative to the one that pushed it (with its payloads) -> (m_{payload}/m_{full}).The root at ~.875 is in the physically obvious place, where it denotes your craft needing to be all engines without fuel to fit in so many stages so close, the vertical asymptote at the origin also makes sense. Finally the maximum occures in a place consitent with the previous graph: .365^4.58 = .0099, which is within expected limits of the craft mass ratio (.001) we declared in part 2 though this equation doesn't rely on knowing specifics about the craft, just the stages. In words, the stage-mass-ratio to the number of stages power equals craft-mass-ratio. Of course you can't have 4.5 stages IRL, but the math doesn't know that .

4. Plot in terms of more convenient (m_{full }/m_{payload}) ... where things are probably definitely wrongFinally, we'll flip (inverse) our dependent variable to make the plot more directly useful when building a rocket from the pod up to the full craft.

define X = (m

_{full }/m_{payload}) = 1/(m_{payload}/m_{full}) . It's more convenient because you normally start with the reentry craft, and want to know how big each subsequent stage to push it should be. Plugging this in to the final equation of part 3 gives what should be the final result:ΔV

=

_{craft}- g_{0 * }I_{sp }_{* }ln(^{1}/_{9}+^{8}/_{9}*TWR_{min}/TWR_{eng}+^{8}/_{9 * }1/(m_{full }/m_{payload}) ) (also simplifying the lower log by pulling out the inverse -> - )ln(m

_{craft}/m_{pod}) ln(m_{full }/m_{payload})But look at that non-zero limit on the right of the graph for huge stages relative to payloads. It should be divebombing, because you get asymptotic deltaV returns with huge stages while the mass increases ever linearly, but the graph seems to be approaching some constant.

Y axis: DeltaV for each e-fold scale of the craft in m/s -> ΔV

_{craft}/ ln(m_{pod}/m_{craft}) ... X axis: relative size of each stage (with its payloads) relative to the one it pushes (with its payloads) -> (m_{full }/m_{payload})With a stage-to-stage mass ratio of 50, we should have an abysmal deltaV efficiency, but the plot claims otherwise. Frustratingly, the maximum is right where we would expect it to be at 2.78 = 1/.36 (.36 being the optimum from before). Given the graphs are all seemingly consistent with eachother, this hints that the actual problem may have shown up a long time ago. Or perhaps not?

So, I hit this road block on Friday and attacked the problem from every angle I could think of over the weekend, rederiving the first 3 sections from different conceptual starting points. I slept on it, mulled over it and triple checked everything, but there's obviously something amiss because the graph in section 4 is obviously wrong in the limit of stages being very large compared to their payloads. Can you figure out the right answer? Best of luck!

Terms(m

_{pod}/m_{craft}) Mass ratio of the entire craft.(m

_{empty}/m_{full}) Mass ratio of this stage when it's empty vs when it's full-> the standard element for the rocket equation.(m

_{payload}/m_{full}) Mass ratio of just this stage's payload to its full stating mass. In other words, the size of each stage (with its payloads) relative to the one that pushed it (with its payloads)As an aside, this happens to be the same value as the size of each stage

(without payload)to the one that pushed it(without payload)when stages scale geometrically as they do here. Not important, just handy.TWR

_{min}/TWR_{eng }The ratio of how much TWR your craft needs at the start of this stage, vs the TWR of your liquid engine by itself. It's a proxy for the mass of the engines relative to the stage.m

_{empty }The mass of this stage when it's empty (no fuel left in the tanks)m

_{full }The mass of this stage when it's full.The rest should be obvious by name or by context, but please let me know if any are confusing and I'll append them here!

Future worksThis can be used to make a lovely graph/plot which will finally simplify the TWR/Isp tradeoff of engines! If I ever get to the bottom of it...

Finally, I'm flagging friendly neighborhood mod @Vanamonde in case another board would be more appropriate. I think this is the most fitting place, but let's see where we end up!

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