Vertical Ascent vs. To LXO First

[GoSlash27]

I'm not sure what the heck he was saying, other than he clearly doesn't like me (not an unusual circumstance )

My entire point (and you may as well get in on this) is that a vehicle that departs the Mun eastward with sufficient DV to return to Kerbin can easily be made for $1,688 (not bothering to drop the cost for 5 solar panels), carrying 45 units of fuel, weighing 1.9 tonnes, and capable of achieving 1,290 m/sec DV can be easily created just by using a Mk.1 command pod, an FL-100 tank, an LV-1 engine, and a handful of solar panels.

There's no vehicle, no matter what the acceleration, that can get anywhere close to these figures when designed to depart upward, so speaking of the efficiency differences as if they are "minor" is frankly bizarre.

And this craft isn't even optimized to the task. It's got way more DV than the mission calls for and the Mk.1 pod isn't the most efficient option.

The math would be DV=9.82*290* ln(1.39/.890)= 1270 and T/w= 4/(1.39*1.63)= 1.77

The in-flight test is that I slapped it together, hyperedited it to LMO, and was able to land it on the surface with about 1/3 of it's fuel still in the tank as proof-of-concept. That, and the fact that people use this combo *all the time* for this job.

So unless there exists a high-g vertical munar departure vehicle that can get anywhere near any of these numbers in any category.... what do we mean by "marginal" when comparing the efficiency of the two departure modes?

As a practical matter, they're not even on the same planet, let alone in the same ballpark.

*That's* what I'm saying.

Best,

-Slashy

Edited December 17, 2014 by GoSlash27

[arkie87]

I'm not sure what the heck he was saying, other than he clearly doesn't like me (not an unusual circumstance )

My entire point (and you may as well get in on this) is that a vehicle that departs the Mun eastward with sufficient DV to return to Kerbin can easily be made for $1,688 (not bothering to drop the cost for 5 solar panels), carrying 45 units of fuel, weighing 1.9 tonnes, and capable of achieving 1,290 m/sec DV can be easily created just by using a Mk.1 command pod, an FL-100 tank, an LV-1 engine, and a handful of solar panels.

There's no vehicle, no matter what the acceleration, that can get anywhere close to these figures when designed to depart upward, so speaking of the efficiency differences as if they are "minor" is frankly bizarre.

And this craft isn't even optimized to the task. It's got way more DV than the mission calls for and the Mk.1 pod isn't the most efficient option.

The math would be DV=9.82*290* ln(1.39/.890)= 1270 and T/w= 4/(1.39*1.63)= 1.77

The in-flight test is that I slapped it together, hyperedited it to LMO, and was able to land it on the surface with about 1/3 of it's fuel still in the tank as proof-of-concept. That, and the fact that people use this combo *all the time* for this job.

So unless there exists a high-g vertical munar departure vehicle that can get anywhere near any of these numbers in any category.... what do we mean by "marginal" when comparing the efficiency of the two departure modes?

As a practical matter, they're not even on the same planet, let alone in the same ballpark.

*That's* what I'm saying.

Best,

-Slashy

A summary of topics discussed:

For the return from Mun case (which is what this thread is about) we are not discussing optimizing cost. The cost discussion was related to Mr. Horseman's challenge.

In this thread, we are talking about for a given craft with high TWR, which transfer method will require least deltaV to get out of Mun's SoI: vertical or horizontal.

We are all aware of the potential gains to be made to the total deltaV of the craft by reducing TWR i.e. using a lighter engine, but the focus of this discussion is not on getting the most deltaV of the craft by picking the right engine, but of the deltaV required by the different transfer methods for a given craft design (one with high but finite TWR).

That said, LethalDose and I provided theoretical calculations for the deltaV required for the different transfer methods (LethalDose even conducted super-accurate simulations of vertical launch with finite TWR and mass loss), and showed that while horizontal impulse burn is most efficient (it's the lower limit on deltaV), in practice, it is not possible, since there might be terrain to clear and it requires an impulse burn.

To that end, I showed that for infinite TWR, if one climbed only 300 m (by accident or to avoid terrain), it would pay to have just departed vertically. For finite TWR, I used LethalDose's excellent results to calculate the height at which point it pays to depart vertical is reduced-- for a TWR of 6, it was only 1 km, and for TWR 3, it was 7.5 km. Thus, even though horizontal burn might be more efficient for infinite TWR (i.e. impulse burns), in practice, there might be terrain in the way or it might be hard to perform accurately enough to keep apoapsis down low enough in order to keep horizontal burn the most efficient transfer method (let alone the fact that it requires an infinite TWR i.e. impulse burn).

To that end, you and LethalDose began to discuss the effect of finite TWR on horizontal burn. LethalDose provided the equations necessary to calculate the required burn angle, as a function of TWR, to balance gravity such that the craft could move horizontally across the surface. For low TWR (which you mention above is both cheaper and has more overall deltaV), the angle becomes more vertical, such that more fuel is wasted preventing vehicle from falling, and in the limit of TWR = 1, the angle is completely vertical such that all fuel is wasted to prevent the craft from falling, and none goes into accelerating the craft horizontally.

The angle required is theta = arcsin(1/TWR); the wasted energy/sec is TWR(1-cos(theta)). The graph of that function looks like this:

For sqrt(2) TWR, approximately 41% of the fuel spent is fighting gravity vs vertical's 100% (however, fuel spent fighting gravity decreases as horizontal velocity approaches orbital due to centripetal lift). So I dont think the solution is immediately obvious in any way...

And i think that is where we left off...

[LethalDose]

A summary of topics discussed:

For the return from Mun case (which is what this thread is about) we are not discussing optimizing cost. The cost discussion was related to Mr. Horseman's challenge.

In this thread, we are talking about for a given craft with high TWR, which transfer method will require least deltaV to get out of Mun's SoI: vertical or horizontal.

We are all aware of the potential gains to be made to the total deltaV of the craft by reducing TWR i.e. using a lighter engine, but the focus of this discussion is not on getting the most deltaV of the craft by picking the right engine, but of the deltaV required by the different transfer methods for a given craft design (one with high but finite TWR).

That said, LethalDose and I provided theoretical calculations for the deltaV required for the different transfer methods (LethalDose even conducted super-accurate simulations of vertical launch with finite TWR and mass loss), and showed that while horizontal impulse burn is most efficient (it's the lower limit on deltaV), in practice, it is not possible, since there might be terrain to clear and it requires an impulse burn.

To that end, I showed that for infinite TWR, if one climbed only 300 m (by accident or to avoid terrain), it would pay to have just departed vertically. For finite TWR, I used LethalDose's excellent results to calculate the height at which point it pays to depart vertical is reduced-- for a TWR of 6, it was only 1 km, and for TWR 3, it was 7.5 km. Thus, even though horizontal burn might be more efficient for infinite TWR (i.e. impulse burns), in practice, there might be terrain in the way or it might be hard to perform accurately enough to keep apoapsis down low enough in order to keep horizontal burn the most efficient transfer method (let alone the fact that it requires an infinite TWR i.e. impulse burn).

To that end, you and LethalDose began to discuss the effect of finite TWR on horizontal burn. LethalDose provided the equations necessary to calculate the required burn angle, as a function of TWR, to balance gravity such that the craft could move horizontally across the surface. For low TWR (which you mention above is both cheaper and has more overall deltaV), the angle becomes more vertical, such that more fuel is wasted preventing vehicle from falling, and in the limit of TWR = 1, the angle is completely vertical such that all fuel is wasted to prevent the craft from falling, and none goes into accelerating the craft horizontally.

The angle required is theta = arcsin(1/TWR); the wasted energy/sec is TWR(1-cos(theta)). The graph of that function looks like this:

http://i.imgur.com/gh085mV.png

For sqrt(2) TWR, approximately 41% of the fuel spent is fighting gravity vs vertical's 100% (however, fuel spent fighting gravity decreases as horizontal velocity approaches orbital due to centripetal lift). So I dont think the solution is immediately obvious in any way...

And i think that is where we left off...

Well, wow...

Apparently someone was paying attention enough to what I was saying that they were able to explain it to someone one else.

Only place I could disagree that I can find is that you called the angle from horizontal theta, and I called it phi (I try to reserve theta for inclination from the equator or ecliptic). And that's just nomenclature, can't really complain about that.

Well done, arkie. I'm impressed.

I can only hope that one day, maybe, someone will build a munar lander with more than three parts, and will find this information useful.

[arkie87]

Well, wow...

Apparently someone was paying attention enough to what I was saying that they were able to explain it to someone one else.

Only place I could disagree that I can find is that you called the angle from horizontal theta, and I called it phi (I try to reserve theta for inclination from the equator or ecliptic). And that's just nomenclature, can't really complain about that.

Well done, arkie. I'm impressed.

I can only hope that one day, maybe, someone will build a munar lander with more than three parts, and will find this information useful.

HOW DARE YOU NOT CALL AN ANGLE THETA! ARE YOU SOME KIND OF BARBARIAN???? DEATH TO PHI!!!!

[LethalDose]

HOW DARE YOU NOT CALL AN ANGLE THETA!

C...cause that reference we both use calls it phi...

Heck, it uses 'gamma' to describe (from what I can tell) the complement of phi... that's just nuts. (see fig 4.7)

Aaaand I just found a new equation in there that would have been insanely useful a few days ago. Eh, well. Knowing is half the hassle.

[Superfluous J]

The cost of the Mun return vehicle is as relevant as the cost of the Kerbin lifter (which I only bring up as reference! Don't worry!). Which is to say it's extremely relevant because it's the easiest way we can judge two ships' relative efficiencies. If two ships can do the exact same thing but one costs twice as much, you'd use the other one so you can either save money or do more.

Much like with the Kerbin lifter, this boils down to: If you build a ship that can get off of Mun and return to Kerbin by lifting straight up, I can build a ship that can also get back to Kerbin but by burning horizontally, for cheaper. Always.

[arkie87]

C...cause that reference we both use calls it phi...

Heck, it uses 'gamma' to describe (from what I can tell) the complement of phi... that's just nuts. (see fig 4.7)

Aaaand I just found a new equation in there that would have been insanely useful a few days ago. Eh, well. Knowing is half the hassle.

Oh. My bad. I betrayed my own source.... oh well.

Please do share this new equation....

[arkie87]

The cost of the Mun return vehicle is as relevant as the cost of the Kerbin lifter (which I only bring up as reference! Don't worry!). Which is to say it's extremely relevant because it's the easiest way we can judge two ships' relative efficiencies. If two ships can do the exact same thing but one costs twice as much, you'd use the other one so you can either save money or do more.

Much like with the Kerbin lifter, this boils down to: If you build a ship that can get off of Mun and return to Kerbin by lifting straight up, I can build a ship that can also get back to Kerbin but by burning horizontally, for cheaper. Always.

This is an assertion without any proof whatsoever. So even if it is 100% true, you will never convince us unless you expect us to just believe you, which is not science. It's dogma.

Edited December 17, 2014 by arkie87

[The_Rocketeer]

This is an assertion without any proof whatsoever. So even if it is 100% true, you will never convince us unless you expect us to just believe you, which is not science. It's dogma.

You do realise that the reason high TWR is better is because it approaches infinite thrust, i.e. impulse burn? Even if both vehicles from both starts could do this, the one using the gravity slingshot would still do it for less fuel. That's not dogma, it's orbital mechanics.

Edit: I really need to stop letting myself get drawn into this.

Edited December 17, 2014 by The_Rocketeer

[Kesa]

Oh!

I just remembered what escape velocity means. It's the velocity at which mechanichal energy (potential + kinetic) becomes positive, therefore you are on escape trajectory no matter the direction of your velocity. It only depends of altitude.

So if you only want to escape the gravity of a body in a single burn, you can point at any direction, the theoritical DV will be the same.

If you only want to leave SOI, in the case of mun, SOI radius/ body radius is 10 so one can prove that to leave SOI, we need sqrt(9/10) times the dv of an escape burn going straight up, and sqrt(90/101) times that same dv going horizontally (using energy conservation and Kepler's 2nd law), so no noticeable difference.

The only differences between paths will be loss, and are well summed up by arkie several posts above.

[Superfluous J]

This is an assertion without any proof whatsoever. So even if it is 100% true, you will never convince us unless you expect us to just believe you, which is not science. It's dogma.

Which assertion, that cost is a valid way to judge two rockets' efficiencies or that you can always build a rocket that goes to orbit first for cheaper?

Cost efficiency is a very important part of this game (and life in general) so sure, that can be my dogma. And as far as the other is concerned, I'll happily make an orbital lifter to prove my point. I only ask that we give it some beef so the gains are more obvious. Say, lift a full orange tank off of the Mun and land it - still full - on Kerbin*. As soon as we finish the other challenge, make a ship that can do this and then I will too. Bonus: We don't even need to have FAR installed.

I'd add "without entering another SOI after leaving Mun's for the first time" just so this is more about the ship and the lifting strategy than any tricks with multiple gravity assists.

[LethalDose]

Oh. My bad. I betrayed my own source.... oh well.

Please do share this new equation....

Oh, it's just 4.44, which gives a vessel's flight angle directly as a function of the it's position vector. I was looking all over for that.

The cost of the Mun return vehicle is as relevant as the cost of the Kerbin lifter (which I only bring up as reference! Don't worry!). Which is to say it's extremely relevant because it's the easiest way we can judge two ships' relative efficiencies. If two ships can do the exact same thing but one costs twice as much, you'd use the other one so you can either save money or do more.

Much like with the Kerbin lifter, this boils down to: If you build a ship that can get off of Mun and return to Kerbin by lifting straight up, I can build a ship that can also get back to Kerbin but by burning horizontally, for cheaper. Always.

In my view, the issue here isn't which is cheaper, it's how much cheaper is it, especially in reference to how technically challenging and/or time consuming the flight path is. And again, to be clear, I agree horizontal burn directly to escape is monotonically the better solution, that's why I used included it in my data and described it as a "lower bound".

I'll say it again, the horizontal burn is always more efficient. Call it science or dogma, or whatever, who cares?

It's just that sometimes, the efficiency gain is so small, it's meaningless.

When you're coming off the back-side of the Mun (near the twin craters) to get to get an encounter with Kerbin, I could launch, and go into the parking orbit, and circularize and then transfer home, or I could just burn straight up. Which one do I want to do? Well it depends. Maybe I don't have the TWR to do a direct vertical ascent efficiently. Maybe I need to get home fast because batteries are dying.

But at least we have the information to make the decision now.

I frankly don't have a lot of faith that a straight vertical burn from Kerbin to ANYWHERE will be a good choice (other than just getting altitude), but on the Mun, there's evidence that sometimes there's no meaningful difference between them.

The point is, there are situations where the difference between the *perfect* ascent and a *redonkulously fast and easy* ascent are practically meaningless, even though they aren't zero.

A real-world comparison to this would be found in statistics. We almost always use asymptotic estimations of distribution for stats instead of their exact counter parts. On the one hand, you can do the asymptotic calculations on the back of an envelope in 5 minutes and get an answer that's within 1% of being correct virtually all the time (This is like the vertical ascent: quick and dirty, but less efficient/accurate). On the other hand, I can use the exact method to get the exact answer and the calculation will take 8 days to run through (Always the most accurate result).

Which one do you think I'm going to choose? By the logic you've presented above, I should always be chosing the more accurate/efficient one, even though it's a freaking headache. I disagree with that choice.

More realistically, lets say the asymptotic calculations give spurious results in certain situations (they do, btw, you need to know how to find them), but still give pretty good, if not perfect, answers in the other situations.

Which one do I choose? Well, it depends on the situation, now doesn't it?

Just like in the vertical/horizontal burn: It's going to depend on the situation. (Of course, this assumes someone something other than the one-and-only Munar lander design). It's not just black and white.

Edited December 17, 2014 by LethalDose

[The_Rocketeer]

When you're coming off the back-side of the Mun (near the twin craters) to get to get an encounter with Kerbin, I could launch, and go into the parking orbit, and circularize and then transfer home, or I could just burn straight up. Which one do I want to do? Well it depends. Maybe I don't have the TWR to do a direct vertical ascent efficiently. Maybe I need to get home fast because batteries are dying.

Wait a minute... is this really what this is about? I thought this was a question of launching straight up from the trailing edge, or horizontally from the leading edge.

If I've got the wrong end of the stick, that explains a lot. For a Mun return to Kerbin, you want to accelerate in a direction retrograde relative to your direction of orbit of Kerbin - that's straight up from near the twin craters. If you were in this situation, there is literally no reason to accelerate horizontally. All the energy you would expend to orbit Mun once is complete wasted.

I thought we were comparing a vertical ascent from the trailing edge of Mun (twin craters) with a horizontal launch from the leading edge of Mun (near farside crater). This is completely different.

[LethalDose]

Wait a minute... is this really what this is about? I thought this was a question of launching straight up from the trailing edge, or horizontally from the leading edge.

If I've got the wrong end of the stick, that explains a lot. For a Mun return to Kerbin, you want to accelerate in a direction retrograde relative to your direction of orbit of Kerbin - that's straight up from near the twin craters. If you were in this situation, there is literally no reason to accelerate horizontally. All the energy you would expend to orbit Mun once is complete wasted.

I thought we were comparing a vertical ascent from the trailing edge of Mun (twin craters) with a horizontal launch from the leading edge of Mun (near farside crater). This is completely different.

This is *roughly* the point I've been trying to make, but without any discussion of specific locations. Whether or not it's what the OP intended when he started the thread, I don't know. From the OP I read this:

Nevertheless, Kerbin isn't the only planet. My question is if you are on the Mun such that Kerbin is perpetually setting (since the Mun is tidally locked) i.e. on the rear-side of the planet w.r.t. Mun's velocity vector, do you want to get into LMO first or go straight vertical to return to Kerbin.

(Emphasis mine) and the passage after it when I gave the first response, then it went all over the freaking place, then I thought it was coming back to Mun departures, so I put together the simulation and reported the results, here.

You may have been in one of the other discussion somehere, and been spot on, but I really have no idea at this point.

What I presented in the linked post was a very general version of what you described above: Burning to escape, regardless of the location. I wanted to get some actual data into the conversation, when all that was happening was a lot of opinions and people telling each other that they're wrong without any evidence to support statements.

I'm not quite sure why we'd compare launch efficiencies from different locations, though...

Edited December 17, 2014 by LethalDose

[arkie87]

Which assertion, that cost is a valid way to judge two rockets' efficiencies or that you can always build a rocket that goes to orbit first for cheaper?

Cost efficiency is a very important part of this game (and life in general) so sure, that can be my dogma. And as far as the other is concerned, I'll happily make an orbital lifter to prove my point. I only ask that we give it some beef so the gains are more obvious. Say, lift a full orange tank off of the Mun and land it - still full - on Kerbin*. As soon as we finish the other challenge, make a ship that can do this and then I will too. Bonus: We don't even need to have FAR installed.

I'd add "without entering another SOI after leaving Mun's for the first time" just so this is more about the ship and the lifting strategy than any tricks with multiple gravity assists.

Sorry, even though i quoted you, what i was referring to was a bit vague... This should clear it up...

Much like with the Kerbin lifter, this boils down to: If you build a ship that can get off of Mun and return to Kerbin by lifting straight up, I can build a ship that can also get back to Kerbin but by burning horizontally, for cheaper. Always.

This is an assertion without any proof whatsoever. So even if it is 100% true, you will never convince us unless you expect us to just believe you, which is not science. It's dogma.

You do realise that the reason high TWR is better is because it approaches infinite thrust, i.e. impulse burn? Even if both vehicles from both starts could do this, the one using the gravity slingshot would still do it for less fuel. That's not dogma, it's orbital mechanics.

Edit: I really need to stop letting myself get drawn into this.

This is dogma. I am asking you to prove it, not just assert it. If you cannot prove it, then dont let yourself get drawn into it, since you cannot help.

[arkie87]

Which assertion, that cost is a valid way to judge two rockets' efficiencies or that you can always build a rocket that goes to orbit first for cheaper?

Cost efficiency is a very important part of this game (and life in general) so sure, that can be my dogma.

Please see my response above. You misunderstood what dogma i was referring to....(though i do not understand why you would think cost was the dogma?)

And as far as the other is concerned, I'll happily make an orbital lifter to prove my point. I only ask that we give it some beef so the gains are more obvious. Say, lift a full orange tank off of the Mun and land it - still full - on Kerbin*. As soon as we finish the other challenge, make a ship that can do this and then I will too. Bonus: We don't even need to have FAR installed.

I'm still working our first challenge...but i think doing a test from an airless body would be an interesting test since I seem to be hitting terminal velocity in FAR... which severely limits benefits of TWR....

I'd add "without entering another SOI after leaving Mun's for the first time" just so this is more about the ship and the lifting strategy than any tricks with multiple gravity assists.

wait-- did you use a gravity assist to get cost below 10k$? If so, i would like to forbid that as well, since gravity assists can be performed using both methods... i.e. you cannot leave Mun's SoI once you enter it...

[The_Rocketeer]

Well, I've conceptually explained at length the reason and process of why and how it works. However, I'm not a mathematician to put it into a nice little equation for you, nor someone who gets a thrill from making up nice little powerpoint presentations to put on youtube and show the world "what I done".

I don't dogmatically believe this is the case, I accept this to be true because my own empirical experience tells me it is. If you don't believe me, try it out for yourself. This isn't dogma, it's traditional knowledge.

I'm not convinced anybody can help you. You decline to demonstrate your own arguments and suggestions with in-game examples time and again, and constantly rely on 'simulation' and mathematical models provided mostly by other people, yet you never categorically prove your own point (whatever it is) because you never test it for yourself.

[arkie87]

Wait a minute... is this really what this is about? I thought this was a question of launching straight up from the trailing edge, or horizontally from the leading edge.

If I've got the wrong end of the stick, that explains a lot. For a Mun return to Kerbin, you want to accelerate in a direction retrograde relative to your direction of orbit of Kerbin - that's straight up from near the twin craters. If you were in this situation, there is literally no reason to accelerate horizontally. All the energy you would expend to orbit Mun once is complete wasted.

I thought we were comparing a vertical ascent from the trailing edge of Mun (twin craters) with a horizontal launch from the leading edge of Mun (near farside crater). This is completely different.

I assume by horizontal launch from the leading edge of the Mun, you are referring to a horizontal burn to LMO that goes directly into the burn to leave Mun SoI (i.e. the lower limit that LethalDose is referring to).

I think we all agree this is most efficient, but like LethalDose said, my question is about a craft on the trailing edge, so the "lower limit--horizontal from the leading edge" is not available since the Mun is tidally locked (I can start a new thread for non-tidally locked moons ).

Also, Like what LethalDose said, the question isnt necessarily which is more efficient, but whether the difference is worth it, given the pros and cons of each method.

[The_Rocketeer]

More to the point, the focus of this thread changes so much and so frequently that it's impossible to keep up a discussion of your original hypothesis. Every time the discussion seems to come back to the point, you (Arkie) take a post by someone else to pieces and fire the debate off on 5 or 6 tangents. The conversation stagnates into pointless bickering because nobody has any idea what anybody else is talking about.

I'm out of this one. I wish you luck in finding your all-encompassing proof - you'll almost certainly need it.

[Superfluous J]

Please see my response above. You misunderstood what dogma i was referring to....(though i do not understand why you would think cost was the dogma?)

I was stabbing in the dark

wait-- did you use a gravity assist to get cost below 10k$? If so, i would like to forbid that as well, since gravity assists can be performed using both methods... i.e. you cannot leave Mun's SoI once you enter it...

We can add that, no prob. FTR I didn't use it, and am not sure how much benefit it would be considering you have to burn up to Mun's SOI before you can get one and by then you're basically where you want to be for the encounter. Coming back to Kerbin, though, you could just burn enough to get out of Mun's SOI and tweak yourself into a Kerbin collision landing with very little further expenditure. Which is against the spirit of this discussion as it would confuse the core issue of orbit vs straight up.

[The_Rocketeer]

I assume by horizontal launch from the leading edge of the Mun, you are referring to a horizontal burn to LMO that goes directly into the burn to leave Mun SoI (i.e. the lower limit that LethalDose is referring to).

I think we all agree this is most efficient, but like LethalDose said, my question is about a craft on the trailing edge, so the "lower limit--horizontal from the leading edge" is not available since the Mun is tidally locked (I can start a new thread for non-tidally locked moons ).

Also, Like what LethalDose said, the question isnt necessarily which is more efficient, but whether the difference is worth it, given the pros and cons of each method.

No, there is literally no reason to reach LMO (or LKO). The periapsis should ideally remain exactly on the surface at the launch site.

Also, how are you going to measure the worth of the difference? If you have excess fuel, it's not worth it. If you have enough fuel for one and not the other, it couldn't be more worth it.

This is a completely pointless discussion. I'm out (for good this time).

[arkie87]

You decline to demonstrate your own arguments and suggestions with in-game examples time and again, and constantly rely on 'simulation' and mathematical models provided mostly by other people, yet you never categorically prove your own point (whatever it is) because you never test it for yourself.

I honestly dont know how you can say this? Have you even read this thread? I have done plenty of my own simulations, and used the equations that LethalDose has used in my OP. I have also done plenty of tests making videos when asked (though not for this case yet since i have been busy doing others).

I agree that experiments can usually trump theory, since theory makes simplifications (such as impulse burns), but experiments also have severe handicaps: while it can take me 5 min to simulate 1000 different cases to see which one is best and/or understand its behavior, it would take me days to fly all of them. And even if i flew all of them, i am still relying on my ability to perfectly execute all of them. Maybe an "ideal burn" (whatever that is for any given case) would be much more efficient, but in reality, if i make a slight mistake, it will actually require the same deltaV as a "less efficient" maneuver, but one that is much easier to execute and/or much less difficult to screw-up. This is the question i am investigating...

And if i provide a mathematical proof why X is better for a given case (which i have in OP), then i expect a mathematical proof of why its not. If you only "proof" is that its not because "orbital mechanics says so", then that doesnt help anyone understand why. I am interested in and trying to figure out the "why", and if you cannot provide that to me, then please stop telling me the "what".

[arkie87]

No, there is literally no reason to reach LMO (or LKO). The periapsis should ideally remain exactly on the surface at the launch site.

By definition, if you burn horizontally up to an altitude, you pass through a circular orbit (assuming impulse burns) with periapsis on the ground. This is what i meant by LMO....

Also, how are you going to measure the worth of the difference? If you have excess fuel, it's not worth it. If you have enough fuel for one and not the other, it couldn't be more worth it.

If both have roughly the same fuel left and cost roughly the same amount, then each player can decide which method is better for them, by weighing the pros and cons. Whatever they decide, more information is better than less information.

This is a completely pointless discussion. I'm out (for good this time).

If you think this discussion is pointless, please bow out, and stop derailing this discussion (after all, lots of other people in this thread derailed the discussion, and then you blamed me for the fact that you were confused about what exactly we are debating).

[GoSlash27]

Wait a minute... is this really what this is about? I thought this was a question of launching straight up from the trailing edge, or horizontally from the leading edge.

If I've got the wrong end of the stick, that explains a lot. For a Mun return to Kerbin, you want to accelerate in a direction retrograde relative to your direction of orbit of Kerbin - that's straight up from near the twin craters. If you were in this situation, there is literally no reason to accelerate horizontally. All the energy you would expend to orbit Mun once is complete wasted.

I thought we were comparing a vertical ascent from the trailing edge of Mun (twin craters) with a horizontal launch from the leading edge of Mun (near farside crater). This is completely different.

Rocketeer,

Actually, it's still more efficient to launch horizontally from the trailing edge. Doing so will put you in an apoapsis with near-orbital velocity overhead the leading edge, with only a small burn necessary to escape munar SOI. This is still a cheaper proposition than burning vertically from the trailing edge.

Were one to increase the t/w enough, I think eventually you could find a point where vertical launch is competitive. Reduced eccentricity due to having been in near-orbit and a small loss in Oberth effect for having drifted to the apoapsis do create a small, but tangible loss in efficiency. I don't believe that they are large enough to overcome the 9m/sec advantage from having launched with the surface rotation, but I might be mistaken on that point.

Assuming I *am* mistaken, then it is theoretically possible that, with enough thrust, one could design a vertical launch vehicle that could edge out the horizontal mode total DV... But that point is moot because nobody would ever design such an inefficient vehicle, drag it's bloated carcass to the mun, and wrestle it down to the surface.

Best,

-Slashy

Edited December 17, 2014 by GoSlash27

[GoSlash27]

A summary of topics discussed:

For the return from Mun case (which is what this thread is about) we are not discussing optimizing cost. The cost discussion was related to Mr. Horseman's challenge.

In this thread, we are talking about for a given craft with high TWR, which transfer method will require least deltaV to get out of Mun's SoI: vertical or horizontal.

We are all aware of the potential gains to be made to the total deltaV of the craft by reducing TWR i.e. using a lighter engine, but the focus of this discussion is not on getting the most deltaV of the craft by picking the right engine, but of the deltaV required by the different transfer methods for a given craft design (one with high but finite TWR).

That said, LethalDose and I provided theoretical calculations for the deltaV required for the different transfer methods (LethalDose even conducted super-accurate simulations of vertical launch with finite TWR and mass loss), and showed that while horizontal impulse burn is most efficient (it's the lower limit on deltaV), in practice, it is not possible, since there might be terrain to clear and it requires an impulse burn.

To that end, I showed that for infinite TWR, if one climbed only 300 m (by accident or to avoid terrain), it would pay to have just departed vertically. For finite TWR, I used LethalDose's excellent results to calculate the height at which point it pays to depart vertical is reduced-- for a TWR of 6, it was only 1 km, and for TWR 3, it was 7.5 km. Thus, even though horizontal burn might be more efficient for infinite TWR (i.e. impulse burns), in practice, there might be terrain in the way or it might be hard to perform accurately enough to keep apoapsis down low enough in order to keep horizontal burn the most efficient transfer method (let alone the fact that it requires an infinite TWR i.e. impulse burn).

To that end, you and LethalDose began to discuss the effect of finite TWR on horizontal burn. LethalDose provided the equations necessary to calculate the required burn angle, as a function of TWR, to balance gravity such that the craft could move horizontally across the surface. For low TWR (which you mention above is both cheaper and has more overall deltaV), the angle becomes more vertical, such that more fuel is wasted preventing vehicle from falling, and in the limit of TWR = 1, the angle is completely vertical such that all fuel is wasted to prevent the craft from falling, and none goes into accelerating the craft horizontally.

The angle required is theta = arcsin(1/TWR); the wasted energy/sec is TWR(1-cos(theta)). The graph of that function looks like this:

http://i.imgur.com/gh085mV.png

For sqrt(2) TWR, approximately 41% of the fuel spent is fighting gravity vs vertical's 100% (however, fuel spent fighting gravity decreases as horizontal velocity approaches orbital due to centripetal lift). So I dont think the solution is immediately obvious in any way...

And i think that is where we left off...

Yessir, I've got that... but all these other factors are critical to the comparison between the two modes. I understand you think you've yourself a favor by eliminating the characteristics of the vehicle itself from the analysis, but in doing so you have eliminated all *practical* vehicles from consideration.

Your analysis, mathematically rigorous as it is, boils down to this:

"If stranded on a desert island, there may be some case where there's a more efficient way to open a coconut than banging it on a rock. You could, for instance, bring an elephant along with you. The math shows that the larger the elephant, the more efficiently it can open a coconut."

I think the reason this gives so many people so much consternation (myself included) is that you speak of metaphorically "bringing along an elephant" and efficiency as if they are not mutually- exclusive concepts.

Stepping back out of the analogy...

Your analysis completely stacks the deck in favor of vertical mode by assuming a vehicle optimized for it while rejecting a vehicle optimized for horizontal launch and erroneously concludes that "there's not much difference between the two modes".

You cannot compare the relative merits of the two modes fairly without comparing *all* of the merits. Vertical launch is easier, while horizontal mode saves DV, weight, cost, and fuel. The difference between the two modes *in total* is not minor in any of these categories.

Vertical launch from the munar surface is not the preferred option *under any circumstances* from a mission planning, engineering, or fiscal standpoint.

As for LD's analysis, the holes in his thinking are threefold:

1)

fuel spent fighting gravity decreases as horizontal velocity approaches orbital due to centripetal lift

^This

2) The t/w of the vehicle increases over time as fuel is consumed.

and

3) The pilot is going to vary the pitch of the vehicle over the duration of the launch, not keep it at a fixed angle.

Best,

-Slashy

Edited December 17, 2014 by GoSlash27