Orbit or Direct Escape Trajectory?

[GoSlash27]

Actually, most of the delta-v gain comes from not fighting gravity as directly.

It's all trigonometry

With a TWR of 2, going directly up, you provide twice as much thrust as needed to balance weight

so you only accelerate at 1g. Half your fuel is directly lost to gravity.

At 45deg from vertical, you've got a twr of 2 along the hypotenuse.

this works out to an acceleration of sqrt(2) horizontally, and (sqrt(2) - 1) vertically

you gain altitude much more slowly, but lose much less energy to fighting gravity.

Since orbit is very little to do with altitude, and almost all about horizontal velocity relative to the surface

this works out more efficient.

As TWR goes to inifinty, the losses to gravity become proportionally less and less important.

Edit to add: For getting into orbit, altitude is required for exactly two things. Avoiding mountains, and avoiding atmosphere.

On a perfectly spherical world with no atmosphere, you could enter orbit at any nonzero ground seperation.

of course, there are no perfectly spherical worlds, so a few KM above highest point is usually a good idea.

Surprisingly, this doesn't work out quite like that in practice. The reduction in DV from higher t/w will be offset by the additional engine mass (and fuel to move that mass, etc) needed to get the t/w and you end up with worse mass efficiency and higher fuel consumption even though you've expended less DV.

Also, the finesse required to get the most out of the t/w becomes increasingly difficult to attain as t/w increases, and the reduction in DV gets lost in the noise floor of human error.

Long/ short, there is an optimal t/w ratio for each engine type and each airless body. It's generally in the 1.4- 2.2 range.

I did a lot of work in this area about a year ago and my findings were validated with another independent study. I'll see if I can find links.

*edit* Here's my results. I will have to redo them, as these apply to the .90 version.

http://forum.kerbalspaceprogram.com/threads/105422-Ideal-T-W-ratios-for-airless-body-launch-(KSP-90)

Here's Tavert's independent findings:

http://forum.kerbalspaceprogram.com/threads/39812-Landing-and-Takeoff-Delta-V-vs-TWR-and-specific-impulse?p=747317&viewfull=1#post747317

Best,

-Slashy

Edited July 16, 2015 by GoSlash27

[soulsource]

I second what Lelitu wrote. Gravity losses should not be underestimated. I once calculated for a return from the Mün, that a direct ascent is only more efficient if the TWR of the ship is above 58. Nevertheless I don't dare to give any numbers for Kerbin, as the calculation I used for the Mün (basically: kick ship a little up, and then accelerate towards the horizon...) definitely doesn't work on bodies with an atmosphere (where one wants to go up for some time to reduce drag losses). Still, once one breaks free from the atmosphere, the trigonometry argument is valid again.

[Weywot8]

Optimizing for dV is like optimizing for "distance on one tank of fuel" that car companies sometimes do in their commercials. It sounds better until you realize that they just gave their gas guzzler a bigger fuel tank.

This insufficient car<->rocket analogy was brought to you by Rockomax. Rockomax: If it ain't broke, it hasn't been field tested enough!

Ah, ok. Good point! I do this in career cause SRB's are cheap for the first stage over tank+fuel+rocket engine and I try to minimize cost and delta-V to orbit over multiple stages.

[PB666]

Newb question here. If I want to travel to Duna or Eve is it better to start from a stabilized orbit of Kerbin or is it better to just take off and keep the nose up until you reach escape trajectory?

To gather the vector you want, essential you want to extend vertical velocity vector at kerbin sunrise. But kerbin is also traveling 170 m/s at its surface, if escape velocity is 3500 m/s then you need to launch 2.78 degrees before sunrise. If the kerbin day is 6 hours long, about 3 minutes before sunrise. The launch needs to be timed such that it is at the Hohmann transfer time for Duna. Since a kerbin year is 400 days long your perfect transfer will be off by +/- 0.3 degrees, in addition since KSP is equitorial you will always have an orbit that is planar with unless you manage to get a launch very close to when Duna orbit crosses the kerbin defined equitorial plane of the Kerbol system, and on that you need to know the pitch north (for a descending node) and south (for an ascending node) and also correct this for the spin of Kerbin.

So here are the problems.

Essentially you are paying for each moment trying to escape after you would have made your gravity turn, the cost is thrust - gravity(altitude). If beyond 45,000 meters your craft has really great acceleraton, say 5g then this is not a problem, you will not spend many moments pushing your weight up, but if it has low acceleration say 1.2 to 1.8 g, the reach orbit, no questions asked, you should choose LKO first.

Second issue is if you plan your trip carefully you can make gravity assists off the mun or minmus, this will be hard to time if you make an escape orbit from kerbin

The third issue you need to know is that to get and land at duna you essentially need low thrust high ISP engines for the transfer, and the place to fire them is at LKO periapsis, the lower the better. Since Duna is so far away and the change of separation angle between the two planets is so slow, you can fire your engines at this periapsis several times for a minute or so each pass to 'kick' the craft to a larger major axis (periapsis + 2 planet radius + apoapsis). So instead of using an inefficient engine that has and ISP of 370 you can use and engine(s) with an ISP of 800 or 4200. You could not use these engines in an escape trajectory from kerbin because their thrust is simply too small. The require circumstances were you can make long burns over time. If you decide to use ION drive engines, remember that Duna gets less light, and the retrograde burn may not point you ideally for sunlight, so make sure you have enough panels and anticipate your burn times appreciably in advance, or you might make a nice flyby on the way out of the kerbol system.

Bottom line,your odds are better with an LKO transition, build a orbit stage that gets you between 80 and 100 km, circularlize. The next stage should be high ISP, there are posts that describe the delta theta suited for transfers to each planet. It is better if you match planes as close to duna as possible, since you will be travelling 1000s of meters per second slower.

Anticipate the SOI with ion drives and begin slowing down, this means your transfer orbit is likely intercept Dunas SOI at a high altitude, and you will begin to retro relative to Duna several days away and adjust your speed downward so that you have a decent chance of making orbit, and essentially reverse the process at duna you performed at kerbin LKO.

Edited July 16, 2015 by PB666

[stelarfox]

what you say is true but only in part.

i have done all the math and, depends truly too much on the ship itself. because it depends completely on the twr

also you are not considering that in order to get in orbit you need to waste 2200 dv between 80 or 70km, if you do not do this you fall.

so if your rocket has a twr of 2, yes you expend a lot more fuel that if you where in orbit but, doing all the math from the point of achieving orbit,

if you actually make orbit (with no fuel loss), you need 2200 dv + 860, meanwhile if with a twr of 2 you just go directly it takes 860*2 (860 for the trust, 860 for the loose, and this is not totaly acurate, you loos that at start you start to loose less as far as you get away (not much but it counts). so, this do cross over.

if your twr = 1 of course you waste all your fuel and stay there (doing it directly up). so of course for a twr of 1 will be more efective to make the orbit first (in fact the only way).

now for a TWR of 1.5, your real aceleration is 0.5 so how much you really need is based on this formula, minDv*TWR/(TWR-1)

this will be exactly the same when this goes to 2200+860 (as you have to expend the rest anyway)  and in the case of the mun mindv =860 so, TWR for this case is 1.39, so thats the only requirement for a "straigth" up to be more efective, of course its also a matter of distance.

even so with this method you can't just calculate that the needed dv is 860 but as i said before 860*TWR/(TWR-1) . and how much you do this is a matter of cost.

so trully is not so good for 1.4 but, its rather good for 1.5 or higher

(and how hard is to reach that its just a matter of your payload).

now if you do not believe me please test it.

 

[Superfluous J]

2 minutes ago, stelarfox said:

if you actually make orbit (with no fuel loss), you need 2200 dv + 860, meanwhile if with a twr of 2 you just go directly it takes 860*2 (860 for the trust, 860 for the loose, and this is not totaly acurate, you loos that at start you start to loose less as far as you get away (not much but it counts). so, this do cross over.

I'm not sure I understand what you're saying here. 2200m/s is orbital velocity and 860 is about Mun transit, so I'm assuming you mean "if you teleport your ship to 100km and are stationary over the planet, you need 2200m/s to get into orbit and then 860 to get to Mun"

Are you saying that if you teleport the same ship to the same place and just burn up for 1720m/s you'll get a Mun intercept? I don't know if that's true or not (but I'm pretty sure it's not), I'm just asking if that's what you're saying.

And I'm also not sure what this has to do with getting to mun (or anywhere) from Kerbin's surface.

[foamyesque]

A parking orbit makes anything other than a trip to the Mun miles easier to plan at a very minimal deltaV cost, and that's as true now as it was when this thread was posted a year and a half ago

[wumpus]

Why should you go to a parking orbit and not a direct escape?  Easy, parking orbit because you can't create a maneuver node on the launch pad.  If you want to determine the differences between direct "escape" and parking orbits I would suggest testing via munshots: they can be easily planned without maneuver nodes (just use them for fine tuning and adjusting your encounter).  This should let you determine if the delta-v savings are worth it (let alone having to manually time your launch).

[I think this second paragraph belongs in some other thread].

Note that while TWR appears to be confusing, I'm fairly sure you shouldn't throttle engines down to achieve your "ideal" TWR (unless it is over 2.0 or something extreme).  I'd recommend adding kickbacks (the biggest SRB) until you have the TWR you need and leaving them all at full thrust (if you have 5 kickers at 80% thrust, you nearly have to do better to only light 4 on the pad [at full thrust] and then light the last when you stage the 4.  But 2nd stage SRBs so rarely make sense that I suspect you would be better off with simply 4 at max thrust and use liquid rockets for the rest).

Edited December 2, 2016 by wumpus

[AeroGav]

The thing is, with a direct ascent we don't necessarily mean "climb vertically off the pad and remain locked on radial out" because that's obviously worst case from a drag loss POV.   Surely it just means a normal launch with gravity turn, towards the end of which you are more or less thrusting horizontally to the horizon.    The only difference is that instead of cutting power and coasting up to AP for a circularization burn, you just keep pedal to the metal till you horizontal velocity exceeds 3000 m/s and AP goes out of Kerbin SOI.    The lower you are the greater the Oberth, so surely completing your burn to escape velocity below 70km is preferable to above, if you can time it right.

Low TWR spaceplane example at the beginning of this vid.   Firewall the throttle and don't stop till we're on our way to Duna..

 

[Tyko]

You can't go from a suborbital path to an escape trajectory without passing through a stable orbit path. At some point on your burn you'll go from a negative PE to a 70Km or higher PE. When you surpass 70Km you're already on an orbital path. You may spend a little DV to fine tune a circular orbit, but that's negligible. Once you're in a circular orbit you have plenty of time to plan your transfer burn and being able to optimize that burn can offer significant DV savings. 

[Superfluous J]

40 minutes ago, Tyko said:

You can't go from a suborbital path to an escape trajectory without passing through a stable orbit path.

Not true. If you burn straight up your Pe will never leave Kerbin's surface, little alone its atmosphere. If you burn perfectly sideways at an Ap of 65km, you can also escape Kerbin's SOI without ever raising the Pe out of the atmosphere.

Under normal play, of course, you are correct. But it's still possible to do and even in normal play I've done it.

[Chiron0224]

On ‎7‎/‎14‎/‎2015 at 11:21 PM, SAI Peregrinus said:

The way I do it (interplanetary missions) is launch to LKO, then to low Minmus orbit where I dock with my fuel station and fill up the rocket, then go interplanetary from there. It makes the initial launch lighter and the fuel comes from the Minmus mining outpost, which also saves Spesos in career.

It took me a second but I realized that you were combining "Space" with "Pesos".  Freaking brilliant.

[Tyko]

3 hours ago, 5thHorseman said:

Not true. If you burn straight up your Pe will never leave Kerbin's surface, little alone its atmosphere. If you burn perfectly sideways at an Ap of 65km, you can also escape Kerbin's SOI without ever raising the Pe out of the atmosphere.

Under normal play, of course, you are correct. But it's still possible to do and even in normal play I've done it.

Fair point. I thought this particular thread was specifically debating circularization versus a gravity ascent followed by direct injection to a transfer vector. I was addressing that, but it's good to clarify 

[Snark]

On 11/30/2016 at 8:40 AM, AeroGav said:

The lower you are the greater the Oberth, so surely completing your burn to escape velocity below 70km is preferable to above, if you can time it right.

Yes, technically you're right.

However, in practice it doesn't make much difference.  Suppose you do a direct ascent where you're effectively ejecting from a 40 km "orbit" instead of an 80 km orbit.  Circular orbital velocity at 40km is 2349 m/s, as opposed to 2278 m/s at 80 km-- just 70 m/s difference.  That seems like an awfully small benefit, given the navigational perks you get from circularizing first.

[GoSlash27]

On 12/1/2016 at 11:21 PM, Snark said:

Yes, technically you're right.

However, in practice it doesn't make much difference.  Suppose you do a direct ascent where you're effectively ejecting from a 40 km "orbit" instead of an 80 km orbit.  Circular orbital velocity at 40km is 2349 m/s, as opposed to 2278 m/s at 80 km-- just 70 m/s difference.  That seems like an awfully small benefit, given the navigational perks you get from circularizing first.

This, and the additional benefit of orbital assembly/ refueling and the question of the gate orbit.

 Theoretically, shooting directly for an interplanetary transfer will save a small amount of DV, but as a *practical* matter... Any DV savings will be erased by the additional correction burns. The most efficient way to do it is to use the vessel as it's own upper stage. Place it in orbit at the gate altitude and refuel it there. Then work out your transfer burn. In many cases, it's actually cheaper in the long run to start from low Munar orbit!

Ultimately, the goal isn't to achieve an intercept with the minimum expenditure of DV, but rather to achieve the intercept with a minimum expenditure of cash.

Best,
-Slashy

 

[stelarfox]

well for mun its truly easy to make it practically better if you know when to make your lunch, a couple of times i manage to make it "so" exactly that i ended at the max mun altitude (refueling there as i put my station to go out of kerbin) and mun is better only because its in the  same plane (no need to adjust like minmus) although its rather possible to do it to minmus but a lot more dificult (at least to do it without correction).

besides, in my personal experience is not always the best to use the cheapest solution, if you use more fuel not a lot but jsut a bit, all the rest you can keep in the space station, instead of making special missions just for that.

 

but then thats not the question here, the question was whats better if make the full orbit, or to go directly and as i stated from the start depends on the ship (i did all the math) even if thats just in theory and in the practice you may expend more due to errors (but you can do the same errors when trying to make your orbit at first too).

and really to just reach mun if you know the angle its far easier to just aim up.

this is sometimes even used to change the plane orbit of real satelites (even so its not done agains gravity).

now sometthing funy is the following, you may never "Make" orbit, you can arrive mun using fly by stop there and get out of it with enough velocity you to fly by too.

in fact i Throwed things from low moon orbit to kerbin (having a ship coupled and just stopping the coupling, then landing that ship again at the moon), the funy thing is that the ship managed to land before the capsule arrived kerbin, and the capsule actually survived all the travel without any issue. (except you cannot delay the landing).

 

[Loren Pechtel]

I must disagree on the reasons for the horizontal burn.  Yes, burning horizontal as much as possible is the way to go but nobody has hit on why!

Consider what happens as you add horizontal velocity:  Centrifugal force is pushing your rocket outward, countering some of the gravity loss.  At orbital velocity 100% of your gravity losses are countered.

Every second you must expend delta-v equal to local gravity/(1 - horizontal velocity/orbital velocity) to keep from crashing, this energy is simply lost.  Obviously, you minimize this loss by building your horizontal velocity as fast as possible.  Until your velocity reaches orbital velocity you should burn vertically only to clear terrain or atmosphere.

Oberth also says you should burn as much as possible as low as possible, again favoring tipping over as soon as you can.

[Sharpy]

Direct launch into optimal trajectory is *slightly* more fuel-conservative, but much harder to perform right. Circularization - bringing your periapsis above the atmosphere - is a few dozen of m/s wasted.

But the truth is this is KSP, not NASA, where weeks of calculations on supercomputers optimize the launch moment and trajectory. In KSP the 'safe time' in LKO when you can prepare an optimal transfer burn will save you more fuel on that transfer, than you'd save on not circularizing but waste on capture burn with your ejection burn being haphazard.

[Red Iron Crown]

7 hours ago, Loren Pechtel said:

Consider what happens as you add horizontal velocity:  Centrifugal force is pushing your rocket outward, countering some of the gravity loss.  At orbital velocity 100% of your gravity losses are countered.

Centrifugal force has nothing to do with it (and doesn't really exist in the relevant reference frame anyway). Burning opposite the direction of gravity's force causes gravity losses, whether you are in orbit or not. A purely horizontal burn on the runway suffers no gravity losses, a radial out one while in orbit does. 

[adsii1970]

To speak in plain English, I'll explain what I do. I've noticed it is more fuel efficient to go into orbit, then use orbital velocity plus an increase in thrust to escape Kerbin's orbit. It is more fuel efficient than a direct escape trajectory approach because you can use gravity as an asset rather than to expend a lot of fuel to fight against it.

 

 

[wumpus]

3 hours ago, Sharpy said:

Direct launch into optimal trajectory is *slightly* more fuel-conservative, but much harder to perform right. Circularization - bringing your periapsis above the atmosphere - is a few dozen of m/s wasted.

But the truth is this is KSP, not NASA, where weeks of calculations on supercomputers optimize the launch moment and trajectory. In KSP the 'safe time' in LKO when you can prepare an optimal transfer burn will save you more fuel on that transfer, than you'd save on not circularizing but waste on capture burn with your ejection burn being haphazard.

Using the first KSP planner hit (and one I like): https://alexmoon.github.io/ksp/

Plotting the first optimal trip to Duna yields: Year 1, day 236 at 4:19:12

Any guesses if launching at exactly 4:19 on year 1, day 236 will provide a sufficiently accurate angle for direct escape?  In practice, I suspect that burning halfway to Mun (with something like 700m/s delta-v) a day earlier would be more effective.  It would take [at most] a tiny amount of circularization (and it would take tremendous thrust to require any) and let you set up a maneuver node for a more accurate flight.

Which brings up the third option for those willing to go beyond KSP's simple tools: direct escape or Mangalayan maneuver?  Manganlayan wins every time if you are using LV-N engines (and simply can't be beat by direct accent with high thrust engines), but I guess it all comes down to which gives you more Obereth (and the "correct" answer is likely a combination of the two, using direct accent to put your AP high enough that the costs in raising PE enough to use the Mangalayan strategy is even possible.

In the end, it comes down to how much abuse you are willing to take avoiding KSP's internal tools for some fairly small delta-v gains.

[Sharpy]

26 minutes ago, wumpus said:

Using the first KSP planner hit (and one I like): https://alexmoon.github.io/ksp/

Plotting the first optimal trip to Duna yields: Year 1, day 236 at 4:19:12

Any guesses if launching at exactly 4:19 on year 1, day 236 will provide a sufficiently accurate angle for direct escape?

 

Absolutely not. Alexmoon's planner absolutely doesn't account for location of KSP. It just says a craft in the 100km orbit, at such phase angle of the orbit that the prograde burn of value given yields the ejection angle given will get the optimal transfer for that situation.

Right at that moment your craft (be it in KSC or in orbit) may be located on the opposite side of Kerbin, pointing in exactly opposite direction, and the burn of the same value will bring it into proximity of Eve orbit instead of Duna.

The times given in the transfer planner must be treated as orientational: +/- 15 minutes of LKO orbital period necessary to get into the right phase angle. Normally, you carefully slide the node (and as result, departure time!) over the orbit until you get the right ejection angle. You can't really do it if you're going on four mammoths roaring from KSP and the trajectory under your node keeps shifting every second!

Quote

burning halfway to Mun (with something like 700m/s delta-v) a day earlier would be more effective

It wouldn't. You'd lose most of benefits of Oberth effect burning from LKO.

The initial orbit will be a compromise between accuracy (thrust at departure burn) and Oberth effect. You lose efficiency with higher orbit, but you lose precision with a lower orbit, as your burn is not a "point burn" but distributed over a longer segment of a more curved trajectory (and waste for a corrective burn to fix the error later).

So, if your escape burn is of order of a minute or less, do it from 70.01km. If it's something closer to 4 minutes, you're better off doing it from some 300-500km. And you'll still need to waste some 100m/s on corrective burn.

Quote

Which brings up the third option for those willing to go beyond KSP's simple tools: direct escape or Mangalayan maneuver?

For engines that won't provide enough thrust to give you an accurate burn, yes, Mangalayan maneuver (multiple periapsis burns repeatedly raising apoapsis) would provide the optimal effect, with full gain of Oberth effect and pretty good accuracy (although the actual departure burn needs to be performed in one go - and it will still be pretty long! But can you plan it? No KSP tools provide the data. And especially that the last burns will bring your orbital period into range of weeks - so how are you going to plan it so that your last periapsis pass, with the actual departure burn, falls into the perfect moment? (actually, using KER, you can - first ballpark the maximum orbital period, then perform the last few burn not to "full extent" but making the orbital period such that you'll arrive at the periapsis at the right time... lots of planning, but doable.)

Edited December 7, 2016 by Sharpy

[Superfluous J]

4 hours ago, adsii1970 said:

To speak in plain English, I'll explain what I do. I've noticed it is more fuel efficient to go into orbit, then use orbital velocity plus an increase in thrust to escape Kerbin's orbit. It is more fuel efficient than a direct escape trajectory approach because you can use gravity as an asset rather than to expend a lot of fuel to fight against it.

 

 

That is better than going straight up, but a mathematical argument can be made that burning to escape on an otherwise gravity-turned orbital ascent saves a scant few m/s over circularizing first.

...and then you lose all that savings doing a huge mid-course correction because you didn't make a nice perfect maneuver node in LKO.

[Tyko]

1 hour ago, Sharpy said:

So, if your escape burn is of order of a minute or less, do it from 70.01km. If it's something closer to 4 minutes, you're better off doing it from some 300-500km. And you'll still need to waste some 100m/s on corrective burn.

This is really useful info. My KSC is primarily building nuke engined craft and the burn times are getting excessive. My solution so far has been to do several kicker burns of about 4 minutes each on successive orbits. With well timed burns I've been getting my actual fuel expenditure to about 10% higher than the projected DV from Transfer Window Planner. This is far better than doing a single 12 minute burn from 100Km in which I'm seeing losses over 25%.

My recent interplanetary burns have cost around 2000DV and have had wastage on the order of 200DV (compared to TWP targets). I guess the question would be if that 200DV would be better spent raising my departure orbit and then just doing one long burn from higher up. It would certainly be easier for navigation planning 

Any suggestions?

Edited December 7, 2016 by Tyko

[wumpus]

1 hour ago, Sharpy said:

Absolutely not. Alexmoon's planner absolutely doesn't account for location of KSP. It just says a craft in the 100km orbit, at such phase angle of the orbit that the prograde burn of value given yields the ejection angle given will get the optimal transfer for that situation.

Right at that moment your craft (be it in KSC or in orbit) may be located on the opposite side of Kerbin, pointing in exactly opposite direction, and the burn of the same value will bring it into proximity of Eve orbit instead of Duna.

The times given in the transfer planner must be treated as orientational: +/- 15 minutes of LKO orbital period necessary to get into the right phase angle. Normally, you carefully slide the node (and as result, departure time!) over the orbit until you get the right ejection angle. You can't really do it if you're going on four mammoths roaring from KSP and the trajectory under your node keeps shifting every second!

That's annoying.  I suppose I have to figure out the angle I need and find the angle I get from a Munshot (the easiest direct escape to compute) and make adjustments to determine what time of day to launch.  It doesn't seem to be included in the information given (because you can simply use the maneuver nodes).  It also requires parsing a bunch of savegame data and looks like a project that would be quickly abandoned.

1 hour ago, Sharpy said:

It wouldn't. You'd lose most of benefits of Oberth effect burning from LKO.

This assumes you are in a circularized orbit, not the highly elliptical one you would get from a direct ascent flight.  Of course you have all the issues of direct escape (and the issues computing the exact time you need to launch) so that your initial orbit is eccentric in the correct path you want to go.  And you also get all the issues of Mangalayan and the dreadfully slow nature of flight.  But you get the advantages of both as well.

You would likely gain even more Oberth this way.  In fact, the nature of orbital flight means that if you burn into and orbit and begin to coast, you will return and cross your path at some point during the burn.  If you were burning full throttle (which of course you would) during your "direct launch", the only way a second burn could lose Oberth would be if you managed to gain *all* your velocity at an AP lower than the atmosphere in a full direct escape launch.  It might happen, but it seems pretty rare.  And the amount of circularization is pretty small (stop when PE is 70k.  You could even stop when 65k, I think it might work better for one burn, and I think Apollo used a very low parking orbit as well), especially when out halfway to the Mun.

I'm pretty sure Mangalayan is the clear winner, but that you should continue burning all the way up to the point you breach the atmosphere (at least).  Of course, that only works for a given rocket with a known TWR.  Which TWR is optimal makes this a vastly more open ended question (aside from the obvious "full thrust from all available engines as the atmosphere thins*"), the choice of engines is the difficult task (presumably with Mangalayan the minimum-fuel choice is either LV-N or LV-909, but players who want to eventually get somewhere might have multiple poodles).

* I'm fairly convinced that you want full thrust much lower as well, but the number of exceptions should tend to zero as the atmosphere thins to zero.