Why Landers?

[tater]

Reality check. Slap legs on your CSM. Land on Mun---but never leave IVA. Let us know how that works out for you.

Pro version: turn on scatter, and if you land and even clip a rock, treat your landing as a catastrophic (all dead) failure.

[mikegarrison]

Reality check. Slap legs on your CSM. Land on Mun---but never leave IVA. Let us know how that works out for you.

You know, IRL we had design options for things like that. A periscope would have made a workable Apollo CSM landing vision aide. These days we would just slap a few video cameras out there and have a better view than through a window anyway.

[tater]

A periscope could be a pretty narrow field of view, or it might be to fish-eyed and distorted. It would be possible to design one such that control inputs would make sense to the pilot, though. Visibility was certainly a factor in LEM design (our lander pods could use a roof window for docking, though).

In addition, if SAS was't magically strong, many slopes would be fatal as well (hard to determine in a periscope or camera view (periscope could be binocular, which would help).

Regardless, even in the tiny kerbol system, landers make some sense even if not actually required.

I would add that the forces on the pilot vs what they see in a periscope system can be very confusing. The USAAF experimented with periscope-controlled gun turrets in WW2, and it made the gunners very quickly nauseous, because the attitude of themselves and what they experienced visually were entirely disconnected.

Edited June 18, 2015 by tater

[Prepper-Jack]

Landers are quite efficient if you bring a large transfer stage into a low polar orbit, and use it as a temporary fuel depot for multiple landings.

Let's say you can land, take off, orbit, and rendezvous from Minmus with less than 180 fuel on your light lander. The big red tank you left in orbit is somewhat spent from getting you there, but still has around 1800 fuel left. That's 9 landings and a return. There's some extra cost associated with getting the extra mass up there - but it's certainly not 9 times as much. If you're good at docking, or at least rendezvous and a claw, it's worth it.

[tater]

This is another place where "realism" would have made gameplay better. In a scaled up Kerbol system (need not be 10X, or even 6.4X, maybe 3-4X), 2-stage landers for the Mun might actually become attractive/necessary. Minmus could then allow for easy, 1 stage landers (or direct descent/ascent) because it might remain small. This would make more variability in "ideal" munar landing operations, which is fun (kerbin-orbit rendezvous, a direct mission, munar-orbit rendezvous, etc). The 2-moon system would be far more interesting if the Mun were substantially larger/difficult. As-is, Minmus is just an ugly, and even easier target.

[AbacusWizard]

Interesting. I'm new to the topic and this thread is gold.

So, what about leaving the fuel in orbit around the Mun, landing with the strictly necessary, and getting back to it later on the way home?

That way one won't need two engines (for lander and mother ship), but just one. What's left in orbit is the fuel for the trip back home.

I've seen (and I think tried once or twice) designs like that as well--the entire ship consists of a small lander with enough internal fuel to take just the lander from Mun orbit down to the surface and back, plus a large external fuel tank with enough fuel to take the whole shebang from low Kerbin orbit to low Mun orbit and back. They're connected by a docking port, with the lander essentially pushing the external fuel tank. Seems to work rather well, especially for a small lander in which saving the mass of that extra engine really makes a difference.

- - - Updated - - -

Another reason for landers that I don't think has been mentioned yet: thrust-to-weight ratio (TWR).

Consider a large interplanetary mothership with a small lander docked to it. The mothership's engines should be chosen based primarily on efficiency; out in space there's (usually) no need to have much thrust. The lander, on the other hand, MUST have a (local) TWR greater than one if catastrophic lithobraking is to be avoided! It is of course easier to get a high TWR on a smaller craft--put small engines on a small lander rather than building a landable mothership with both efficient engines and enormous landing engines.

This is even more important for long-term multi-target missions. If you're just going to Minmus and back, sure, make an all-in-one craft that can get there, land, and get back; if on the other hand you're planning on visiting both Duna and Ike in one trip, or all the moons of Jool, or even setting up a permanent Minmus science base, it makes far more sense to leave the excess fuel (and interplanetary engines, science lab, crew quarters, and anything else that won't be needed on the surface) in orbit with a docking port.

[mikegarrison]

Pre 1.xx I used to go for a standard "lander" design that was really more of an all-in-one ship. I would stick the 3-person pod on top of a MPL lab, put a clampotron sr. on the bottom, and have four widely spaced radial fuel/engine/leg stacks for thrust and landing stability. For Mun I would just have a disposable transfer stage to get it there, and would usually fly directly home after without a rendezvous. For outer planets I would have an interplanetary tanker stage that I would leave in orbit when I landed, then dock to and use for the trip home. The tanker would have very low TWR ratio but could carry a lot of fuel with a high ISP. The lander would have lower ISPs but enough TWR to work almost anywhere.

In 1.xx, however, it's harder and more expensive to lift something like that out of KSC. So I've been trying more minimalistic and aerodynamic landers. As the physics model changes, so do the designs that respond to them.

Edited June 18, 2015 by mikegarrison

[MaverickSawyer]

I've always utilized a multi-ship system when in Sandbox. (Career, depends on how far along I am...) I'll often design an insanely efficient transfer vehicle for crews, and have them meet a dedicated lander at the target body. Often, I'll also leave the lander there for another mission to use and pack along the extra fuel in the transfer ship. But now, with stock ISRU, I simply add an unmanned refinery ship to the mission and refuel on the surface. That way, the lander is actually carrying fuel for a round trip from surface to surface. Not the easiest way of doing things, but it works.

[Arsonik]

If RSS could be integrated into stock in such a way that the difficulty sliders on career/science proportionally effected planet scale and distance.... Well..... I can dream, right?

In 1.xx, however, it's harder and more expensive to lift something like that out of KSC. So I've been trying more minimalistic and aerodynamic landers. As the physics model changes, so do the designs that respond to them.

Have you considered just cutting your old design in half, or separating the radial sections off? Just use a stack separator for the command pod instead of docking port (just to reduce the number of docking ports, doesn't really matter), and add the fuel/engine/leg stacks via docking? Or just launch the standard version empty and refill on a second launch? Using the vertical snap you can get docking ports perfectly centered on the fuel stacks /center stack by pressing "V" while mousing over the docking ports and after attaching them to the stack in question. ((I think "V" haven't played in a few days)(Might only be with Editor Tools?)) This will make building the docking ports for radial stacks symmetrically very easy (literally the click of a button). Sure you're looking at 4 launches now for 4 stacks, but you can replace those stacks and extend the re-usability of the design in case an engine breaks or something.

Heck, for say a Joolian mission you could just leave the lander there for return secondary/tertiary missions later on. Just package the 4 radial parts docked to a couple of large structural beams or something to keep it compact. You could probably fit 4 2.5m stacks into one (albeit large) fairing that way. Send it all out in one go to your lander, with crew/return vehicle of course. Repeat for additional missions ad nauseam.

I may try building something along these lines myself incorporating the science lab into the primary lander. Usually I keep a station in orbit with a lab for most bodies and do multiple landings as necessary. Won't have to rendezvous as much this way, and I can probably get a ton of dV out of some orange tank stacks.

Edited June 18, 2015 by Arsonik

[Eric S]

I don't think that's true.

If you want to get into an orbit, then yes, you want to burn sideways and generate that velocity vector. But if you just want to get out of the Mun's gravity well, it seems like going straight up would be best. Why generate a velocity vector sideways that you aren't going to use?

Except that you are using it, and as was said, gravity losses. I also suspect that thrusting straight up doesn't benefit as much from the Oberth effect.

Think of the gravity losses this way. Say you have a craft with a 2.0 TWR. This means that when you burn straight up, you lose half your thrust fighting gravity. On the other hand, if you burn at 60 degrees below vertical (30 above the horizon), you wind up with a vertical TWR of 1.0 and a horizontal TWR of 1.732. Gravity cancels out the vertical TWR, leaving you accelerating horizontally 73% faster than you were accelerating vertically. Then, as you gain horizontal velocity, the amount of vertical TWR you lose to gravity becomes even less because you're coming closer to being in orbit.

Now, that first factor is scaled by the Pythagorean triangle type math, so the higher your TWR, the less important it becomes. At a TWR of 3, you're only accelerating horizontally 1.414 times faster than you would if you burned straight up. However, diminishing returns kick in and the horizontal acceleration is still higher until you have an infinite TWR.

As for the first part (you are using it), that has to do with orbital mechanics. The specific energy of an orbit never changes, though unless you're in a circular orbit you're constantly trading between potential energy and kinetic energy. Basically, if your orbit's specific energy is higher than the potential energy at SoI altitude plus the kinetic energy of an orbit at that altitude, your orbit will leave the SoI (barring hitting atmosphere or terrain, or a delta-v burn) and your speed as you reach the SoI is determined strictly by how much more orbital energy you have than potential enery at that height (speed after crossing the SoI depends on the actual vector plus the body's vector relative to the parent body). If what matters is the orbital energy, then the heading of your velocity doesn't actually matter for purposes of exiting the SoI.

That second part ignores the direction you're leaving, but that can be adjusted easily enough by proper selection of when you do your transfer burn.

There is a corner case where an orbit with more energy than the potential energy at SoI altitude but less than that potential energy plus the kinetic energy of an orbit at that altitude may or may not exit the SoI, but that's usually not particularly significant unless your intention is to barely leave the SoI. If you're leaving with enough velocity to transfer somewhere, you tend to need enough extra energy that this is a non-issue, and if you're barely leaving the SoI then doing a transfer burn to go somewhere else, you're spending more delta-v than is necessary to get there.

So, the most efficient way to leave the Mun and return to Kerbin isn't to burn straight up unless you've got an infinite TWR. On the other hand, while you do want horizontal velocity, it is probably more efficient to burn to SoI exit rather than doing an actual circularization burn, though to be honest, the difference between burning to orbit and burning directly to transfer are probably relatively minor.

TLDR: gravity losses are real so you accelerate faster if you don't burn directly against gravity, and horizontal velocity is just as good as vertical velocity for leaving an SoI at any significant velocity.

[Laie]

Except that you are using it, and as was said, gravity losses. I also suspect that thrusting straight up doesn't benefit as much from the Oberth effect.

I did some experiments a while ago. Task was to get into a Minmus-like orbit. It turned out that it barely mattered, dV-wise, whether you did a classical launch into orbit or if you just went straight up and took a sharp left at the end. If anything, the straight-up approach seemd to be ever so slightly cheaper.

This came unexpected, to say the least.

The straight-up solution performed better or worse depending on it's TWR, but I had to give it very low TWRs in order to make it look bad. For any reasonable amount of thrust, the final dV expenditure was pretty similar.

Of course, this were Kerbin ascents, where straight-up takes the fastest route out of the atmosphere. But still: If your task is to leave a SOI (so you don't have to circularize at all), going straight up may be the cheaper solution. Not by much, and it will be pretty difficult to aim it... but it certainly won't waste a lot of fuel and probably will save a little.

Rationale (may be wrong): if you want to leave, you can take the direct route and don't need to raise the periapsis one bit.

[mikegarrison]

If your task is to leave a SOI (so you don't have to circularize at all), going straight up may be the cheaper solution. Not by much, and it will be pretty difficult to aim it... but it certainly won't waste a lot of fuel and probably will save a little.

Rationale (may be wrong): if you want to leave, you can take the direct route and don't need to raise the periapsis one bit.

That was my point.

Let's look at the Mun. In Kerbin coordinates, your Mun lander is already in orbit around Kerbin while it is sitting on the Mun surface. The most efficient way to drop back down is to burn retrograde, right? But if you are sitting on the Mun surface and you happen to be pointed retrograde to your Kerbin orbit, burning retrograde is the same thing as launching straight up from the Mun.

- - - Updated - - -

Except that you are using it, and as was said, gravity losses. I also suspect that thrusting straight up doesn't benefit as much from the Oberth effect.

Think of the gravity losses this way. Say you have a craft with a 2.0 TWR. This means that when you burn straight up, you lose half your thrust fighting gravity. On the other hand, if you burn at 60 degrees below vertical (30 above the horizon), you wind up with a vertical TWR of 1.0 and a horizontal TWR of 1.732. Gravity cancels out the vertical TWR, leaving you accelerating horizontally 73% faster than you were accelerating vertically. Then, as you gain horizontal velocity, the amount of vertical TWR you lose to gravity becomes even less because you're coming closer to being in orbit.

Yes, yes, but WE AREN'T GOING TO ORBIT. The most efficient way to simply leave the gravity well is straight up, max thrust. All this Pythagoras stuff is about developing a sideways velocity vector so that you can orbit. If you never intend to orbit, that's just wasted delta-v.

[Archgeek]

Yes, yes, but WE AREN'T GOING TO ORBIT. The most efficient way to simply leave the gravity well is straight up, max thrust. All this Pythagoras stuff is about developing a sideways velocity vector so that you can orbit. If you never intend to orbit, that's just wasted delta-v.

Not quite, and as said before, you're actually wasting delta-v going straight up. The cheapest way home from the mun is hard and flat from little more than 90 degrees shy of Kerbin retrograde. This lets you whip around the Mun with an ejection angle pointing fully retrograde to Kerbin, barely fighting the Mun's gravity at all, and getting that sweet sweet Oberth goodness from doing the whole transfer so deep in the Mun's gravity well. Do the same thing from just shy of Kerbin retrograde (to make up for Munar rotation), and burn straight up, and you've got the same nice ejection, the same Oberth party, but you're leaking delta-v by directly fighting the Mun's gravity.

[magnemoe]

I've seen (and I think tried once or twice) designs like that as well--the entire ship consists of a small lander with enough internal fuel to take just the lander from Mun orbit down to the surface and back, plus a large external fuel tank with enough fuel to take the whole shebang from low Kerbin orbit to low Mun orbit and back. They're connected by a docking port, with the lander essentially pushing the external fuel tank. Seems to work rather well, especially for a small lander in which saving the mass of that extra engine really makes a difference.

- - - Updated - - -

Another reason for landers that I don't think has been mentioned yet: thrust-to-weight ratio (TWR).

Consider a large interplanetary mothership with a small lander docked to it. The mothership's engines should be chosen based primarily on efficiency; out in space there's (usually) no need to have much thrust. The lander, on the other hand, MUST have a (local) TWR greater than one if catastrophic lithobraking is to be avoided! It is of course easier to get a high TWR on a smaller craft--put small engines on a small lander rather than building a landable mothership with both efficient engines and enormous landing engines.

This is even more important for long-term multi-target missions. If you're just going to Minmus and back, sure, make an all-in-one craft that can get there, land, and get back; if on the other hand you're planning on visiting both Duna and Ike in one trip, or all the moons of Jool, or even setting up a permanent Minmus science base, it makes far more sense to leave the excess fuel (and interplanetary engines, science lab, crew quarters, and anything else that won't be needed on the surface) in orbit with a docking port.

I have used the idea of having extra fuel in the transfer stage who is left in orbit so I can dock, refuel and do an secondary landing.

Another common tactic is to use one engine from circulating to doing the braking burn from 10 m/s to 5 back on Kerbin.

You use drop tanks. Then have 3-4 radial tanks who expands the base of your lander it also hold goo and material labs, legs and extra batteries. this is dropped at takeoff or then run dry.

For interplanetary missions you also need decent living rooms for crew and life support. You don't land this unless your ship also is an base and you use another ship to swap crew.

[mikegarrison]

Not quite, and as said before, you're actually wasting delta-v going straight up. The cheapest way home from the mun is hard and flat from little more than 90 degrees shy of Kerbin retrograde. This lets you whip around the Mun with an ejection angle pointing fully retrograde to Kerbin, barely fighting the Mun's gravity at all, and getting that sweet sweet Oberth goodness from doing the whole transfer so deep in the Mun's gravity well. Do the same thing from just shy of Kerbin retrograde (to make up for Munar rotation), and burn straight up, and you've got the same nice ejection, the same Oberth party, but you're leaking delta-v by directly fighting the Mun's gravity.

I guess I'm going to have to find some time to do the calculus, but I think this is wrong. You have to "fight the Mun's gravity" no matter what.

Let's specify the question. You are already sitting on Mun, facing retrograde to your Kerbin orbit. (Ignore Munar rotation for now, because it is a small effect.) Is there any direction you can burn other than straight up which will give you a lower delta-v return back to Kerbin?

I'm not asking if there is any other location on Mun that has a lower d-v requirement to get back to Kerbin. I'm asking, from that spot (facing retrograde), is there any trajectory that gets you back to Kerbin better than just straight up? I think not.

[Rhomphaia]

Not quite, and as said before, you're actually wasting delta-v going straight up. The cheapest way home from the mun is hard and flat from little more than 90 degrees shy of Kerbin retrograde. This lets you whip around the Mun with an ejection angle pointing fully retrograde to Kerbin, barely fighting the Mun's gravity at all, and getting that sweet sweet Oberth goodness from doing the whole transfer so deep in the Mun's gravity well. Do the same thing from just shy of Kerbin retrograde (to make up for Munar rotation), and burn straight up, and you've got the same nice ejection, the same Oberth party, but you're leaking delta-v by directly fighting the Mun's gravity.

You really don't waste much more.

Escape velocity is escape velocity. If you could make instantaneous burns you would not waste any.

Since you cant, for a vertical ascent you fight gravity for the duration of the burn, the horizontal ascent is only fighting until it gets to orbital velocity, so would be slightly more efficient, but for realistic TWRs the difference is pretty insignificant.

The major drawback to the vertical ascent is that, since Mun is tidally locked, it seriously limits your choice of landing sites where it is possible.

[mikegarrison]

For interplanetary missions you also need decent living rooms for crew and life support. You don't land this unless your ship also is an base and you use another ship to swap crew.

Not in stock KSP you don't. If you are roleplaying, or if you are using a mod that requires it, then OK. But in stock KSP you don't need "living rooms".

[Jouni]

I did a quick experiment with a probe that landed on Mun. With a single 48-7S engine, the initial TWR was 5.94 on the surface. A direct vertical ascent until the escape velocity required 833 m/s, while a horizontal ascent took just 777 m/s.

It seems that there is a small but noticeable difference between horizontal and vertical ascents, even if you're about to leave the SoI before going to orbit first.

Edited June 18, 2015 by Jouni

[Eric S]

I did some experiments a while ago. Task was to get into a Minmus-like orbit. It turned out that it barely mattered, dV-wise, whether you did a classical launch into orbit or if you just went straight up and took a sharp left at the end. If anything, the straight-up approach seemd to be ever so slightly cheaper.

This came unexpected, to say the least.

Leaving an atmosphere is going to add a factor favoring the straight up approach. I've seen people test this and come up with the orbit being slightly more efficient. I suspect that the real difference may be minor enough that it can get lost in minor variations of other factors. Also, I'd expect this trick to be more viable going to Minmus than to the Mun due to it's lower orbital velocity. But that's not what the post was discussing.

Yes, yes, but WE AREN'T GOING TO ORBIT. The most efficient way to simply leave the gravity well is straight up, max thrust. All this Pythagoras stuff is about developing a sideways velocity vector so that you can orbit. If you never intend to orbit, that's just wasted delta-v.

It's not wasted delta-v, as I pointed out in the second paragraph after what you quoted. It's increasing the orbital energy in the most cost-effective manner. You're hung up on the fact that you don't want to go to orbit, but that doesn't mean you aren't still bound by orbital mechanics. If you're not being supported by something else, you're in an orbit, even if you're moving straight up. Show me where my math is wrong, because it's saying that for purposes of escaping an SoI, the actual heading of the velocity rarely matters, the speed is far more important.

That's not saying that I think that going straight up from the Mun is going to be horribly inefficient, it's just going to be less efficient. The Mun's gravity isn't strong enough to make doing things in a less efficient manner that noticeable and also means that your TWR is going to be higher. Doing this from some place without an atmosphere but with a higher gravity, say Tylo, will cost you noticeably more.

Also, I'd like to apologize if I'm coming off as snippy or like a dog that won't let go of a bone. It's not my intention, orbital mechanics are not the most intuitive thing so not understanding stuff like this is reasonable, I'm just in enough physical pain at the moment that my already less than impressive social skills are a bit frayed.

EDIT: On the subject of the original post: I think that a large part of the reason a munar rendezvous isn't as effective as a lunar rendezvous mostly comes down to two factors. First, the fact that the landing, takeoff, and return burn take a lot less delta-v on the Mun than on the Moon. Second, the lower TWR of KSP engines means that there's a lot less to be gained by frequent staging, because you're carrying more inactive engine mass. I once did an Apollo-style Mun mission and had to create custom fuel tanks because the smallest 2.5m fuel tank had much more fuel than I needed for the landing, taking off, and return burn all by itself.

One other possible reason is that the LEM was designed from top to bottom to be as light as possible, I read somewhere that there were parts of the craft where stepping on the craft while it was under full earth gravity would have put your foot through the panel. The Mk 2 lander can, on the other hand, weighs almost as much as the Mk 1-2 lander can on a per-kerbal basis.

Edited June 18, 2015 by Eric S

[maccollo]

I'm not asking if there is any other location on Mun that has a lower d-v requirement to get back to Kerbin. I'm asking, from that spot (facing retrograde), is there any trajectory that gets you back to Kerbin better than just straight up? I think not.

North, south, east, west. It doesn't matter what direction you pick. Any direction will work as long as you leave within an orbit or so, and will be substantially more efficient than going straight up assuming you TWR isn't extremely high.

But why don't you actually try it?

download this quicksave file. Jebediah is on the surface of the moon, facing munar retrograde. Your challenge is to get him and his pod back to Kerbin.

https://dl.dropboxusercontent.com/u/22015656/quicksave.sfs

[Randazzo]

It's usually just an aesthetic choice for me. If we're using the Mun as an example, I'd need to tack on at least an extra 1600 dV to the main ship in order to land with it and return to Kerbin. While it's possible, I don't like funkily designed ships/landers. It's nicer looking to have a separate craft for the job.

[mikegarrison]

Leaving an atmosphere is going to add a factor favoring the straight up approach. I've seen people test this and come up with the orbit being slightly more efficient. I suspect that the real difference may be minor enough that it can get lost in minor variations of other factors. Also, I'd expect this trick to be more viable going to Minmus than to the Mun due to it's lower orbital velocity. But that's not what the post was discussing.

It's not wasted delta-v, as I pointed out in the second paragraph after what you quoted. It's increasing the orbital energy in the most cost-effective manner. You're hung up on the fact that you don't want to go to orbit, but that doesn't mean you aren't still bound by orbital mechanics. If you're not being supported by something else, you're in an orbit, even if you're moving straight up. Show me where my math is wrong, because it's saying that for purposes of escaping an SoI, the actual heading of the velocity rarely matters, the speed is far more important.

That's not saying that I think that going straight up from the Mun is going to be horribly inefficient, it's just going to be less efficient. The Mun's gravity isn't strong enough to make doing things in a less efficient manner that noticeable and also means that your TWR is going to be higher. Doing this from some place without an atmosphere but with a higher gravity, say Tylo, will cost you noticeably more.

Also, I'd like to apologize if I'm coming off as snippy or like a dog that won't let go of a bone. It's not my intention, orbital mechanics are not the most intuitive thing so not understanding stuff like this is reasonable, I'm just in enough physical pain at the moment that my already less than impressive social skills are a bit frayed.

I'm not being terribly social myself. And I think we may be talking past each other about different problems, to some extent.

[Xeldrak]

Considering reality, there is also this problem:

(Comparison of lunar lander sizes, from an early Langley study)

Not only is it impossible to see the landing spot while landing, but the direct ascend "lander" would be about 20 meters high. And these Space-Suits were stiff and bulky.

According to wiki the complete Apollo A7L weighs 91kg. Consider climbing a 20m ladder in such a suit.

[tater]

^^^in KSP people would make ridiculous, wide landers and not worry about the height (or use magical jetpacks instead of a ladder). Heck, they'd then want the entire craft to reenter Kerbin and land, lol

[Orbital Vagabond]

Yes, yes, but WE AREN'T GOING TO ORBIT. The most efficient way to simply leave the gravity well is straight up, max thrust. All this Pythagoras stuff is about developing a sideways velocity vector so that you can orbit. If you never intend to orbit, that's just wasted delta-v.

This completely incorrect from an orbital mechanics perspective. In fact, this is the least efficient way to break out of a gravity well.

Some proprotion of dV spent thrusting along the radial direction (e.g. "straight up") is lost due to gravity acting in opposition to the thrust (I refer to this as gravity drage, but some other people might disagree with that term. It's not important). Thrusting perpendicular to the gravity vector (e.g. a gravity turn) mitigates losses due to gravity.

This can be demonstrated in a simple thought experiment as follows:

Someone up above pointed out that "Escape velocity is escape velocity" (Escape velocity is actually dependent on altitude, but we'll let that slide for the moment). Also, Fuel consumption and thrust are constant over time.

When acceleration is parallel to the gravity vector, acceleration due to thrust is reduced due to gravity, as in a "straight up launch".

When acceleration is pependicular to the gravity vector, acceleration due to thrust is not reduced by gravity, as in thrusting to a parking orbit.

The vessel thrusting "straight up" will have lower acceleration than the vessel thusting perpendicular to gravity. Hence, the a vessel on the "straight-up trajectory" will have to thrust for more time than the vessel on the perpendicular trajectory to reach escape velocity. More time thrusting directly translates to more fuel consumption.

The parking orbit also provides some advantages to the straight up trajectory, in particular you can select your ejection angle; You can't select your ejection angle when launching straight up from a tidally locked body.

I say "from a orbital mechanics" point of view, because there are a few things that can influence empirical results. First, high TWR and high velocities well in excess of escape velocity can mitigate those loses (greater speed leads to less time in the sphere of influence, reducing gravity drag). Very high TWRs are relatively easy to attain on low gravity bodies like the Mun and Minmus. Also, a straight-up trajectory reduces the amount of time the vessel suffers from atmospheric drag, which may influence empirical results when taking off from Kerbin.

Addendum: On this comment...

I'm asking, from that spot (facing retrograde), is there any trajectory that gets you back to Kerbin better than just straight up? I think not.

There's a pair of videos videos demonstrating the difference between these two approaches from the Mun

. The vertical ascent was slightly less efficient than the parking orbit (878.7 m/s compared to 855.3 m/s). Edited June 18, 2015 by Orbital Vagabond