Vertical Ascent vs. To LXO First

[Superfluous J]

As a side point, "prograde" is a bit vague since launching vertically is "prograde" as well. I prefer to say "horizontal" vs. "vertical" rather than "vertical" vs. "prograde".

Not true. On an airless body you should, pre-launch, switch to "Orbit" mode because the moment you lift off the ground "surface" mode is irrelevant. "Prograde" at launch is always directly east, as all bodies in KSP rotate east. Therefore, for a perfect launch you should thrust directly east on takeoff though that's impossible for several reasons, not the least of which being that you will almost assuredly die in a fiery crash.

Regarding the rest, why would you burn vertical 10km and then turn? I frequenly take off from Mun at about a 5-10 degree angle off of horizontal. Works like a charm in 99% of the cases. For those rare instances where a crater wall will ruin my day, I go up a little more.

[arkie87]

Actually I won't argue that. That was an early test and right after posting I instantly realized that 5 SRBs at 60% thrust wastes dV. I redesigned the rocket to use 3 SRBs at 100%, drop them, and then use 2 others at 100%. Almost the same TWR with extra dV and a longer burn time.

(And I'm still curious to see your sub-$10k craft)

Sorry for putting words in your mouth then Though you do admit that you have said that in the past-- but now you realize that it is less efficient. Your solution to stage the SRB's sounds better. So my guess is, my craft will likely look like yours, except while you stage the SRB's and go horiztonal, I will put all 5 on at once, and go vertical...

I am still waiting for FAR to come out for 0.90...

And can you verify cost of your craft has not changed in 0.90 (Squad did some balancing, no? so part costs might have changed).

[arkie87]

Not true. On an airless body you should, pre-launch, switch to "Orbit" mode because the moment you lift off the ground "surface" mode is irrelevant. "Prograde" at launch is always directly east, as all bodies in KSP rotate east.

Really? are you really arguing about this?

Regarding the rest, why would you burn vertical 10km and then turn? I frequenly take off from Mun at about a 5-10 degree angle off of horizontal. Works like a charm in 99% of the cases. For those rare instances where a crater wall will ruin my day, I go up a little more.

I said this in my own post, that 10km is a bit exaggerated, and its better to aim slightly upward if you arent near the crater wall...

I am going to make a graph of at what altitude it pays to keep going vertical rather than turn as a function of TWR.

[GoSlash27]

Arkie,

For such a low TWR, of course, burning vertically will not pay. The discussion must involve high TWR (even if you would have had more deltaV had you chosen a lighter engine and lower TWR).

Aye, but there's the rub: Any rocket that is *designed* to launch East is going to have a light engine and low t/w ratio, because a high t/w and heavy engine is completely useless for such a craft.

Your argument is that a craft, if given a high enough T/W can out-perform another copy of itself by going vertical in this scenario... which could *possibly* be the case if t/w is sufficiently high, but it's a moot argument because any craft designed to go vertical will *by nature* be grossly inefficient compared to a craft designed to go horizontal in every category.

So when you say the differences would be "minor", it would only be true if you happened to have placed a grossly inefficient pig of a rocket on the surface of the mun in the first place.

FAR and stock are identical out of atmosphere, so we can directly compare results. I can build a rocket that will outperform your best vertical effort in every category of efficiency in the scenario you have proposed here. DV, cost, fuel consumption, total vehicle mass... everything. And the difference between these two vehicles wouldn't be anything you could call "minor"..

The scenario is as follows: Place a vehicle of your own design (stock parts, no cheats) in the location you describe carrying a Mk.1 command pod as payload. Build it to be as efficient as you can and post the specs (mass, cost, fuel). Then escape from the Mun vertically and establish a periapsis within Kerbin's atmosphere. Again, post mass and fuel state. You may use solar panels to provide power, since they are massless. You may deduct the cost of the solar panels for the purposes of this exercise. You will not be charged for any parts you use to place the lander on the surface itself or decouplers to detach the launch vehicle from the lander, as placing the payload on the surface is not part of the test.

I will do the same, but going horizontal. I guarantee my rocket will blow your rocket's doors off.

Best,

-Slashy

Edited December 16, 2014 by GoSlash27

[LethalDose]

Arkie,

Aye, but there's the rub: Any rocket that is *designed* to launch East is going to have a light engine and low t/w ratio, because a high t/w and heavy engine is completely useless for such a craft.

Your argument is that a craft, if given a high enough T/W can out-perform another copy of itself by going vertical in this scenario... which could *possibly* be the case if t/w is sufficiently high, but it's a moot argument because any craft designed to go vertical will *by nature* be grossly inefficient compared to a craft designed to go horizontal in every category.

So when you say the differences would be "minor", it would only be true if you happened to have placed a grossly inefficient pig of a rocket on the surface of the mun in the first place.

FAR and stock are identical out of atmosphere, so we can directly compare results. I can build a rocket that will outperform your best vertical effort in every category of efficiency in the scenario you have proposed here. DV, cost, fuel consumption, total vehicle mass... everything. And the difference between these two vehicles wouldn't be anything you could call "minor"..

Best,

-Slashy

This argument isn't going to get you as far as you might think...

Let's try to limit the discussion the situation at hand here, which has been specified as Mun escapes repeatedly. We can expand to generalities later. The difference between launching east and launching west on the Mun is, at worst, ~18 m/s. So the dV cost is kind of a wash.

For lift-off TWR from an airless body, if you want to to accelerate in any direction horizontally, you obviously have to offset the vessel's downward acceleration due to gravity. Given that your craft and generate thrust in a single direction, the angle you have to hold relative to the horizon (Φ) is limited by your vessel's TWR; Specifically: Φ = sin(1/TWR). If you fall below this angle, you'll fall to the ground. You need a TWR of 11.5 to hold a Φ at 5 degrees, and a TWR of 5.8 to hold Φ at 10 degrees.

Now, these TWR values are getting to be riiiiight in the range where the vertical ascent is breaking even with the traditional ascent. You can claim to work wonders with your engineering, but you really can't beat the math.

And the issue isn't engine TWR, it's TWR for the whole vessel, including payload and reaction mass. Since you need similar TWRs and dV's for both ascent methods, engineering is really much less of a factor.

You seriously need to start talking in concrete values and numbers, or give some examples, or there's no way the discussion can proceed productively.

[Superfluous J]

I am still waiting for FAR to come out for 0.90...

And can you verify cost of your craft has not changed in 0.90 (Squad did some balancing, no? so part costs might have changed).

I will wait for you to make a craft that can get the payload into a 30km orbit of Mun before I do any more work on this. I've made a ship without knowing exactly what your ship would be. It's your turn.

[arkie87]

First, I'd like to thank you for your work!

Question: for the two horizontal cases, I assume you are using the standard orbital equations (not performing a simulation) since I have matched the values you have given using these equations. It is worth noting that these values will be slightly worse in the real world due to non-infinite TWR i.e. assuming maneuvers are performed with impulse burns. However, doing these simulations would be complicated and rather annoying...

DIRECT ELLIPTIC ASCENT

The ideal direct assent was calculated assuming an instant, prograde acceleration from sea level that would create an eliptic orbit that would reach the SOI. The dV calculated for such an assent was calculated to be 766.7 m/s, and, IMO, should be considered as an absolute lower bound for a Munar escape from sea level.

I used the equation and got 776 m/s (i assume that's just a typo or maybe rounding error). Escape velocity is 807 m/s, as a reference.

10 KM PARKING ORBIT

My estimated dV to go from sea level to a circular 10 km parking orbit was 575.4 m/s* using a eastward burn from the equator. An additional 198.7 m/s to transfer from the parking orbit to the SOI boundary. The total dV for these maneuvers was 774.1 m/s. The Mun's orbital surface velocity (~ 9 m/s) was included in these calculates.

*This value seems quite low to me, especially since the circularization burn was ~ 7 m/s. A quick glance at dV maps put this value between 580 and 640, based on the altitude of the parking orbit. Assuming 640 m/s is correct for a 14 km parking orbit, the entire escape would take approximately 840 m/s. Even with this upward bump, this is still ~ 12.5% lower than arkie presented in the OP.

It should also be noted the overall dV cost of the parking orbit escape increases as the parking orbit altitude increases.

This is actually a very important point. Parking orbit altitude (regardless of whether it is obtained by vertically climbing or climbing by slingshotting around the planet) can make vertical burn more efficient than horizontal burn. I have repeated your calculations done and presented them in graphical form:

From the graph, direct horizontal burn is always most efficient, but not always practical for two reasons:

(1) You might be below nearby terrain, and therefore, have to climb. Climbing at a shallow angle before velocity is comparable to orbital will not result in centripetal lift and will waste fuel.

(2) You cannot perform an impulse burn, and therefore, must aim slightly upwards, even if you could avoid all terrain.

(3) You are not Mechjeb and will not perform perfectly.

The graph shows that if you climb up 300 m (that's VERY low), turning 90 degrees and burning horizontally and keeping on going vertically will require the same deltaV. However, in the real world, you do not have infinite TWR, and the break-even point will change. For a TWR of 6, from LethalDose's result, we need 830 m/s; that relates to roughly a 1 km climb being the break even point. For a TWR of 3, from LethalDose's result, we need 910 m/s; that relates to roughly a 7.5 km climb being the break even point.

[arkie87]

Arkie,

Aye, but there's the rub: Any rocket that is *designed* to launch East is going to have a light engine and low t/w ratio, because a high t/w and heavy engine is completely useless for such a craft.

Your argument is that a craft, if given a high enough T/W can out-perform another copy of itself by going vertical in this scenario... which could *possibly* be the case if t/w is sufficiently high, but it's a moot argument because any craft designed to go vertical will *by nature* be grossly inefficient compared to a craft designed to go horizontal in every category.

So when you say the differences would be "minor", it would only be true if you happened to have placed a grossly inefficient pig of a rocket on the surface of the mun in the first place.

FAR and stock are identical out of atmosphere, so we can directly compare results. I can build a rocket that will outperform your best vertical effort in every category of efficiency in the scenario you have proposed here. DV, cost, fuel consumption, total vehicle mass... everything. And the difference between these two vehicles wouldn't be anything you could call "minor"..

The scenario is as follows: Place a vehicle of your own design (stock parts, no cheats) in the location you describe carrying a Mk.1 command pod as payload. Build it to be as efficient as you can and post the specs (mass, cost, fuel). Then escape from the Mun vertically and establish a periapsis within Kerbin's atmosphere. Again, post mass and fuel state. You may use solar panels to provide power, since they are massless. You may deduct the cost of the solar panels for the purposes of this exercise. You will not be charged for any parts you use to place the lander on the surface itself or decouplers to detach the launch vehicle from the lander, as placing the payload on the surface is not part of the test.

I will do the same, but going horizontal. I guarantee my rocket will blow your rocket's doors off.

Best,

-Slashy

Many people (including myself) have made this argument already.... I dont know why you keep on bringing it up.

This isnt a design question (yet... see below); it is a hypothetical question of what to do if you find yourself in X scenario.

I actually think the case 5thHorseman suggested is a perfect test of this:

We have a lander craft that must be lifted to orbit around the Mun. This craft is specified it cannot be changed. The challenge is to design the lifter stage(s).

He has 5 SRB's in two stages (3 first, then 2) and will launch horizontally from Kerbin into LKO followed by periaps transfer to Mun; I will have 5 SRB's as well, but in one stage burning vertically. We will both be using the same craft, but with different staging. Same potential TWR, same mass, same cost, same everything just the optimal staging for each burn.

This design by 5thHorseman is the result of trying to minimize cost (<10k$). And we see, that by minimizing cost, we are led to a design with SRB's that has a potentially high TWR. Furthermore, we can manipulate this design (without adding or removing parts--just staging) to have a high TWR-- optimal for vertical ascent-- or low TWR--optimal for horizontal.

[LethalDose]

First, I'd like to thank you for your work!

Question: for the two horizontal cases, I assume you are using the standard orbital equations (not performing a simulation) since I have matched the values you have given using these equations. It is worth noting that these values will be slightly worse in the real world due to non-infinite TWR i.e. assuming maneuvers are performed with impulse burns. However, doing these simulations would be complicated and rather annoying...

I used the equation and got 776 m/s (i assume that's just a typo or maybe rounding error). Escape velocity is 807 m/s, as a reference.

Yes, they're calculated using standard orbital equations, which can be found here and elsewhere.

Two notes on escape velocity. First, Escape velocity is altitude dependent, hence it is impossible to state a single value for escape velocity, e.g. "Escape velocity is 807 m/s, as a reference." You would need to state the altitude at which you must exceed 807 m/s to escape the Mun. Eq 4.78 in the linked resource should make this relationship obvious. Second, while the relationship between escape velocity and altitude remains in patched conics (e.g. higher altitudes require lower Vesc), the escape velocity values in KSP's patches conics are slightly below what the traditional equation yields, since the Vesc IRL is the speed needed to attain a an infinitely high AP, not a set AP for an SOI change.

As stated in my post, I calculated the dV needed for a prograde burn from PE to reach an AP = SOI limit, not the traditional escape velocity.

I also want to be clear on my position in this discussion: that I think that the horizontal burns are the more efficient method to escape orbit. In an absolute sense, I think my previous post demonstrates that the dV costs for the horizontal burns is lower than the costs for vertical burns. In relative sense, though, I don't think there's a big difference between the two given a reasonable TWR. Certainly not a difference that will "blow your rocket's doors off".

Edited December 16, 2014 by LethalDose
Clarification of position.

[GoSlash27]

For lift-off TWR from an airless body, if you want to to accelerate in any direction horizontally, you obviously have to offset the vessel's downward acceleration due to gravity. Given that your craft and generate thrust in a single direction, the angle you have to hold relative to the horizon (Φ) is limited by your vessel's TWR; Specifically: Φ = sin(1/TWR). If you fall below this angle, you'll fall to the ground. You need a TWR of 11.5 to hold a Φ at 5 degrees, and a TWR of 5.8 to hold Φ at 10 degrees.

Now, these TWR values are getting to be riiiiight in the range where the vertical ascent is breaking even with the traditional ascent. You can claim to work wonders with your engineering, but you really can't beat the math.

I really have no idea what you're talking about here.

Launching off of the surface at these angles, your math would hold true, but not in a vertical departure from the surface, which would be any *realistic* scenario.

If the t/w is precisely 1, then at the instant of ignition nothing will happen. But as the fuel burns off, the ship will have t/w > 1, and the excess is used to provide horizontal acceleration IAW Pythagoras.

It's not an "engineering miracle", it's common practice.

The challenge is open to you as well, if you'd care to try your hand at it.

Best,

-Slashy

[arkie87]

Yes, they're calculated using standard orbital equations, which can be found here and elsewhere.

This is my source too

Two notes on escape velocity. First, Escape velocity is altitude dependent, hence it is impossible to state a single value for escape velocity, e.g. "Escape velocity is 807 m/s, as a reference." You would need to state the altitude at which you must exceed 807 m/s to escape the Mun. Eq 4.78 in the linked resource should make this relationship obvious.

Yes, I am aware. Escape velocity was given at sea level. It's a useful sanity check to compare our numbers to.

Second, while the relationship between escape velocity and altitude remains in patched conics (e.g. higher altitudes require lower Vesc), the escape velocity values in KSP's patches conics are slightly below what the traditional equation yields, since the Vesc IRL is the speed needed to attain a an infinitely high AP, not a set AP for an SOI change.

As stated in my post, I calculated the dV needed for a prograde burn from PE to reach an AP = SOI limit, not the traditional escape velocity.

Yes, i was going to point this out, but i think you actually mentioned that in your post

Your absolute minimum burn 775.6 m/s is below escape velocity at sea level, which is only possible since you leave the SoI of Mun. If Mun were not orbiting Kerbin, You would need 807 m/s. (I know you know this, but I am explaining it for others).

[arkie87]

I really have no idea what you're talking about here.

Launching off of the surface at these angles, your math would hold true, but not in a vertical departure from the surface, which would be any *realistic* scenario.

If the t/w is precisely 1, then at the instant of ignition nothing will happen. But as the fuel burns off, the ship will have t/w > 1, and the excess is used to provide horizontal acceleration IAW Pythagoras.

It's not an "engineering miracle", it's common practice.

The challenge is open to you as well, if you'd care to try your hand at it.

Best,

-Slashy

If you burn up with TWR = 1, nothing happens.

If you burn horizontally with TWR = 1, you accelerate sideways while falling down.

If you are at ground level, you cannot burn horizontally at TWR = 1. You need TWR >1, and obviously, the greater the TWR, the greater the angle you can use. And it's obviously most efficient to have a higher angle away from vertical.

Thus, you have to burn at an angle to counteract gravity. He is calculating that angle....

Edited December 16, 2014 by arkie87

[GoSlash27]

Many people (including myself) have made this argument already.... I dont know why you keep on bringing it up.

This isnt a design question (yet... see below); it is a hypothetical question of what to do if you find yourself in X scenario.

I actually think the case 5thHorseman suggested is a perfect test of this:

We have a lander craft that must be lifted to orbit around the Mun. This craft is specified it cannot be changed. The challenge is to design the lifter stage(s).

He has 5 SRB's in two stages (3 first, then 2) and will launch horizontally from Kerbin into LKO followed by periaps transfer to Mun; I will have 5 SRB's as well, but in one stage burning vertically. We will both be using the same craft, but with different staging. Same potential TWR, same mass, same cost, same everything just the optimal staging for each burn.

This design by 5thHorseman is the result of trying to minimize cost (<10k$). And we see, that by minimizing cost, we are led to a design with SRB's that has a potentially high TWR. Furthermore, we can manipulate this design (without adding or removing parts--just staging) to have a high TWR-- optimal for vertical ascent-- or low TWR--optimal for horizontal.

But that's a crap design and nobody's going to find themselves in that scenario! That's *my* point. It goes back to "crappy rocket is crappy no matter which way you point it".

Why would anyone in their right mind design a $10,000 heap with 5 (five!!) SRBs when they can do the job for less than $2,000, 1.4T total mass, and a half ton of liquid fuel?? And that's without really trying.

And what monstrosity are you gonna use just to place that Edsel on the Munar surface? You gotta do the TMI hauling that ginormous payload, and get all of that off of Kerbin.

And you think the differences between these two approaches will be "marginal"?? I'm half- tempted to expand the challenge to a full Munar mission from launch to splashdown and spot you the 1Km/sec+ advantage from running FAR, it's so bad.

What started this whole ballyhoo was the idea that "maybe vertical launch isn't as awful as people think", but as a practical matter, it *is* awful. Not just awful, but truly *horrendous*.

Sorry,

-Slashy

Edited December 17, 2014 by GoSlash27

[LethalDose]

What started this whole ballyhoo was the idea that "maybe vertical launch isn't as awful as people think", but as a practical matter, it *is* awful. Not just awful, but truly *horrendous*.

Sorry,

-Slashy

I really have no idea what you're talking about here.

Launching off of the surface at these angles, your math would hold true, but not in a vertical departure from the surface, which would be any *realistic* scenario.

If the t/w is precisely 1, then at the instant of ignition nothing will happen. But as the fuel burns off, the ship will have t/w > 1, and the excess is used to provide horizontal acceleration IAW Pythagoras.

It's not an "engineering miracle", it's common practice.

The challenge is open to you as well, if you'd care to try your hand at it.

Best,

-Slashy

I've done my work, presented it, and clearly stated anyone not actually bringing data to the table is wasting everyone's time.

Until you've presented something, you're wasting my time.

Go demonstrate your own amazing rocket, and link the craft. Give us an example of a munar lift off from wherever you want, record it, put it on YouTube. Then I'll fly the exact same vessel and get results consistent with what I've presented.

It's on you.

[Superfluous J]

Why would anyone in their right mind design a $10,000 heap with 5 (five!!) SRBs when they can do the job for less than $2,000, 1.4T total mass, and a half ton of liquid fuel?? And that's without really trying.

Several notes:

• $10k is the total cost of the rocket to lift the payload arkie87 specified. Note: That payload costs more than $2,000.
  
• That $10k rocket (actually closer to $7k if you don't incude the payload) is designed to take that payload from the launchpad to a 30km (max) Munar orbit. Perhaps you misunderstood the goal as launching the vessel from the Mun to orbit? It's understandable considering how many different bodies we've discussed launching from.
  
• I would love to see a $2k rocket (Total $5k, let's say, counting the payload) that could get the specified payload into the desired location. It would literally change the way I play the game from now on.
  

[arkie87]

But that's a crap design and nobody's going to find themselves in that scenario! That's *my* point. It goes back to "crappy rocket is crappy no matter which way you point it".

Why would anyone in their right mind design a $10,000 heap with 5 (five!!) SRBs when they can do the job for less than $2,000, 1.4T total mass, and a half ton of liquid fuel?? And that's without really trying.

And what monstrosity are you gonna use just to place that Edsel on the Munar surface? You gotta do the TMI hauling that ginormous payload, and get all of that off of Kerbin.

And you think the differences between these two approaches will be "marginal"?? I'm half- tempted to expand the challenge to a full Munar mission from launch to splashdown and spot you the 1Km/sec+ advantage from running FAR, it's so bad.

What started this whole ballyhoo was the idea that "maybe vertical launch isn't as awful as people think", but as a practical matter, it *is* awful. Not just awful, but truly *horrendous*.

Sorry,

-Slashy

I think you really need to read what I wrote rather than responding so quickly.

The challenge is to take a lander from kerbins surface and get it into orbit around mun. If you can do that for 2000 kerbucks, then I and 5thhorseman will be shocked.

[GoSlash27]

If you burn up with TWR = 1, nothing happens.

If you burn horizontally with TWR = 1, you accelerate sideways while falling down.

If you are at ground level, you cannot burn horizontally at TWR = 1. You need TWR >1, and obviously, the greater the TWR, the greater the angle you can use. And it's obviously most efficient to have a higher angle away from vertical.

Thus, you have to burn at an angle to counteract gravity. He is calculating that angle....

I understand all that, but he's wasting his time.

You are not constrained to maintain a single pitch angle during a launch, and your t/w increases as fuel burns off, so launching at 1G vertically means that you *will* rise from the moment the first molecule of fuel meets the first molecule of oxygen.

You don't need a huge t/w to get off the mun in an eastbound trajectory, and everyone who's ever played KSP already knows this.

Scratchin' mah head,

-Slashy

[LethalDose]

Also... I'm confused by this:

I will do the same, but going horizontal. I guarantee my rocket will blow your rocket's doors off.

I really have no idea what you're talking about here.

Launching off of the surface at these angles, your math would hold true, but not in a vertical departure from the surface, which would be any *realistic* scenario.

You say you're going to launch horizontally, but when I address launching horizontally, you claim vertical departures are the realistic scenario...

Seriously, it's time to amaze us with how wrong we are with some actual facts.

[GoSlash27]

No, the challenge is to get *your* lander from the surface and place it in munar orbit. (Honestly, I thought that it was to get a reentry to Kerbin).

If it's just getting *a* lander, then *OF COURSE* it can be accomplished at my price point, and below that if it's just orbit. Not like it's hard; you'd have to search far and wide to find a lander that someone put on the Mun that was anywhere near $10,000, because nobody would ever over-engineer a lander that badly unless they were just plain awful at engineering.

I'm surprised this shocks you,

-Slashy

[GoSlash27]

Also... I'm confused by this:

You say you're going to launch horizontally, but when I address launching horizontally, you claim vertical departures are the realistic scenario...

Seriously, it's time to amaze us with how wrong we are with some actual facts.

LD,

Have you ever launched a lander off the mun?

If so, did you launch it vertically and pitch eastward like everybody else? I ask this in all seriousness because I can't fathom any other reason why any of this would puzzle you.

Believe me, I'm not assuming that you're clueless. This *HAS* to be a failure in communication.

Truly confused now,

-Slashy

Edited December 17, 2014 by GoSlash27

[arkie87]

No, the challenge is to get *your* lander from the surface and place it in munar orbit. (Honestly, I thought that it was to get a reentry to Kerbin).

If it's just getting *a* lander, then *OF COURSE* it can be accomplished at my price point, and below that if it's just orbit. Not like it's hard; you'd have to search far and wide to find a lander that someone put on the Mun that was anywhere near $10,000, because nobody would ever over-engineer a lander that badly unless they were just plain awful at engineering.

I'm surprised this shocks you,

-Slashy

You really need to read what I wrote.

Lander from surface of kerbin to orbit around mun.

Not surface of mun to kerbin.

I honestly can't tell if you are just a troll...

[GoSlash27]

All,

I have to assume that the failure in communication is on my end.

When I say "Launch vertical", I mean that the rocket departs the surface vertically. When I say "depart eastward", I mean pitch eastward after having left the surface so that the vertical component of thrust balances gravity and the remainder is used to accelerate eastward (clearing terrain as necessary) until apoapsis is achieved.

I *HAVE* to assume that you guys are talking about something else entirely and I'm just not getting it.

[GoSlash27]

Wait,

Scratch everything I've said in the last couple pages.

*Which thread are we in again??*

Dude, this is too many different threads on the same topic.

Is this one about escape from the Munar surface, or escape from Kerbin?

I thought this one was about escaping the Mun.

I think this is the source of the confusion...

-Slashy

[arkie87]

All,

I have to assume that the failure in communication is on my end.

When I say "Launch vertical", I mean that the rocket departs the surface vertically. When I say "depart eastward", I mean pitch eastward after having left the surface so that the vertical component of thrust balances gravity and the remainder is used to accelerate eastward (clearing terrain as necessary) until apoapsis is achieved.

I *HAVE* to assume that you guys are talking about something else entirely and I'm just not getting it.

If you launch with twr of one, the angle above horizontal from eastward is 90 degrees

[GoSlash27]

No, scratch my previous post.

*THIS* thread is about your proposed munar departure.

Check your OP.

Seriously, too many different threads on the same topic and you're confusing yourself along with the rest of us.

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.

Your OP in this thread.

This one ain't on me,

-Slashy