Transfering to Mun: First to LKO or Directly to Mun?

[KerikBalm]

raising your PE doesn't cost any dV that you wouldn't spend to get to the moon, either way you burn at Ap until your PE becomes your Ap, and your Ap extends out to the Mun.

so the circularization burn is not a problem.

Its Oberth you want.... doing the injection burn at a higher velocity... which is at a lower orbit. If this was a perfectly spherical airless body (similar to tylo) you might as well go into orbit first. This is no different than Perapsis Kicking.

Of course, orbital velocity gets higher the lower you go. So you'll get more oberth effect from doing your munar injection from 60km than from 70km.

Of course, then you also get more dV lost to atmospheric drag due to higher speeds within the atmosphere.

I'm not sure at which point the increased drag cancels out the increased oberth effect - but i think the difference is so small, you might as well go to LKO, and then precisely tune your injection manuever node, rather than trying to rush a munar injection when you may not be quite at the optimal window.

For the purposes of recovery, I like to get into orbit, and then separate the craft, and recover the orbital injection stage. You could do this when still suborbital, but then the stage will land far away from the KSC.

[arkie87]

Forget the maths, send me the craft files and tell me what mods you use and I'll fly them both using both methods (straight up vs orbit first). I'll post it on YouTube. All your friend needs to do to refute me is to get his craft to Mun via direct ascent and have more fuel in his tank than I do after getting to orbit first.

I'm using FAR (the only mod relevant to physics) and stock parts. I can send you the craft files if you really want, but they are kinda difficult to fly with a gravity turn (no torque).

I would recommend making your own simple craft with about 7 km/s deltaV -- orange fuel tank + mainsail + reaction wheel + control surfaces and small fuel tank + lv909 and command pod should be close?

If you make a video, mention me for giving you the idea! I want to be famous!!

Edit: Now i kinda want to try this myself.... and make a video too, but that will have to happen later tonight...

[Wanderfound]

raising your PE doesn't cost any dV that you wouldn't spend to get to the moon, either way you burn at Ap until your PE becomes your Ap, and your Ap extends out to the Mun.

so the circularization burn is not a problem.

Eh?

1) Launch spaceplane, build speed and altitude until apoapsis exceeds 70km (periapsis will be somewhere around 30km at this stage).

2) Immediately go into transfer burn.

3) Complete transfer burn, with periapsis still at about 30km.

In a flat-climb spaceplane ascent, your periapsis is often fairly close behind you when your apoapsis breaks out of the atmosphere. If you go straight into a transfer burn (without waiting to coast around to apoapsis to circularise), you'll be burning much closer to Pe than Ap.

[numerobis]

I hear what you are saying, but those effects should be potentially equal depending on the encounter. Regardless, the best comparison we can make is to get both ships into a circular orbit with same altitude as the Mun and compare.

The Oberth effect isn't linear. The Hohmann from 0m altitude gets you a speed 375 m/s on arrival at the Mun SoI; that costs 276 m/s to burn off into a 30km circular orbit. You claim 530 m/s at arrival, which costs 359 m/s. So this is more in favor of vertical ascent.

With gravity assists you can theoretically get that down arbitrarily close to 0 m/s on arrival, which costs 184 m/s to burn off.

I'm calculating that with the planet.soiBurn routine in my KSP-scripts, which I implemented via a thread here:

http://forum.kerbalspaceprogram.com/showthread.php/16511-Interplanetary-How-To-Guide

[arkie87]

The Oberth effect isn't linear. The Hohmann from 0m altitude gets you a speed 375 m/s on arrival at the Mun SoI; that costs 276 m/s to burn off into a 30km circular orbit. You claim 530 m/s at arrival, which costs 359 m/s. So this is more in favor of vertical ascent.

With gravity assists you can theoretically get that down arbitrarily close to 0 m/s on arrival, which costs 184 m/s to burn off.

I'm calculating that with the planet.soiBurn routine in my KSP-scripts, which I implemented via a thread here:

http://forum.kerbalspaceprogram.com/showthread.php/16511-Interplanetary-How-To-Guide

I'm not sure if i follow. I think more words or perhaps mathematical relations would help me understand what you are trying to say.

Vertical ascent gets you to Mun's SoI with 0 m/s horizontal velocity relative to Kerbin, and takes 530 m/s to burn horizontally to get into a circular orbit around Kerbin at Mun's altitude. Not sure if thats what you understand me to mean. If so, then disregard this...

[arkie87]

By the way, I think you missed the 175m/s sidereal rotation you get for free on launch. That's a gift to the standard launch approach, and does almost nothing for the vertical launch.

Yes, i stated that in the assumptions. It helps both, but i think it helps gravity-turn more.

Do you have words to explain what you're trying to calculate with that EES code? It's best to talk about concepts so we can each do the math in a different way and get similar numbers.

The code is commented out, but basically works as follows:

Each paragraph/group contains one approach. Each approach goes through each burn, calculates deltaV's and then sums them up.

For each burn, there are two equations that relate distance 1 and velocity 1 to distance 2 and velocity 2, and those are conservation of energy (the ones involving velocity squared terms) and kepler's (2nd?) law (distance_1*velocity_1 = distance_2*velocity_2).

The only other equation is V^2/d = mu/d^2 which simplifies to V^2 = mu/d and is used to calculate the velocity needed for a circular orbit at distance d.

EDIT: it might also help to know that EES is a numerical solver of systems of non-linear equations. Thus, rather than use a and b to calculate c, i can specify 3 equations with a, b, and c in them, and it will calculate all three at the same time. So, each line doesnt have to calculate something, but rather, give a relationship between variables.

It's a really powerful software. Highly recommended

Edited December 11, 2014 by arkie87

[arkie87]

Yes, i stated that in the assumptions. It helps both, but i think it helps gravity-turn more.

The code is commented out, but basically works as follows:

Each paragraph/group contains one approach. Each approach goes through each burn, calculates deltaV's and then sums them up.

For each burn, there are two equations that relate distance 1 and velocity 1 to distance 2 and velocity 2, and those are conservation of energy (the ones involving velocity squared terms) and kepler's (2nd?) law (distance_1*velocity_1 = distance_2*velocity_2).

The only other equation is V^2/d = mu/d^2 which simplifies to V^2 = mu/d and is used to calculate the velocity needed for a circular orbit at distance d.

EDIT: it might also help to know that EES is a numerical solver of systems of non-linear equations. Thus, rather than use a and b to calculate c, i can specify 3 equations with a, b, and c in them, and it will calculate all three at the same time. So, each line doesnt have to calculate something, but rather, give a relationship between variables.

It's a really powerful software. Highly recommended

More Heavily Commented EES Code:

{!Constants}

{Distances}

R=600e3 "Kerbin Radius"

A=100e3 "LKO Altitude"

H=12000e3 "Height of Desired Orbit--in this case: Mun's"

{Gravitational Parameters}

mu=3.5316e12 [m3/s2] "Kerbin's General Gravitational Parameter: G*M"

{!Vertical Ascent}

{Initial Climb}

-mu/R + 1/2*V[0]^2 = -mu/(R+H) "Conservation of Energy"

{Circularization}

V[1]^2 = mu/(R+H) "Momentum Balance"

{Delta-V's}

dV[0]=V[0]

dV[1]=V[1]

{!Horizontal Ascent to LKO then to Mun}

{Ascent to LKO}

-mu/R + 1/2*V[2]^2 = -mu/(R+A) + 1/2*V[3]^2 "Conservation of Energy"

R*V[2] = (R+A)*V[3] "Kepler's Law"

{Ascent to Mun}

-mu/(R+A) + 1/2*V[4]^2 = -mu/(R+H) + 1/2*V[5]^2 "Conservation of Energy"

(R+A)*V[4] = (R+H)*V[5] "Kepler's Law"

{Circularization}

V[6]^2=mu/(R+H) "Momentum Balance"

{Delta-V's}

dV[2]=V[2]

dV[3]=V[4]-V[3]

dV[4]=V[6]-V[5]

{!Horizontal Ascent from Surface}

{Ascent to Mun}

-mu/R + 1/2*V[7]^2 = -mu/(R+H) + 1/2*V[8]^2 "Conservation of Energy"

R*V[4] = (R+H)*V[8] "Kepler's Law"

{Circularization}

V[9]^2=mu/(R+H) "Momentum Balance"

{Delta-V's}

dV[5]=V[7]

dV[6]=V[9]-V[8]

{!Summing Up Different Approaches}

dV_vertical = dV[0]+dV[1]

dV_horz = dV[2]+dV[3]+dV[4]

dV_horz_surf = dV[5]+dV[6]

[arkie87]

The Oberth effect isn't linear. The Hohmann from 0m altitude gets you a speed 375 m/s on arrival at the Mun SoI; that costs 276 m/s to burn off into a 30km circular orbit. You claim 530 m/s at arrival, which costs 359 m/s. So this is more in favor of vertical ascent.

With gravity assists you can theoretically get that down arbitrarily close to 0 m/s on arrival, which costs 184 m/s to burn off.

I'm calculating that with the planet.soiBurn routine in my KSP-scripts, which I implemented via a thread here:

http://forum.kerbalspaceprogram.com/showthread.php/16511-Interplanetary-How-To-Guide

I think what you are implying is that circulaization step isnt necessary, since a gravity assist can be used to help.

If we dont count circulaization step, then vertical is 4 m/s better with no atmosphere Inside atmosphere, it's probably a few hundred m/s better...

[numerobis]

No, what I'm saying is that the Hohmann transfer's second burn of 375 m/s, or the 540 m/s you use, tell you the speed you have relative to Mun when you enter its SoI. That speed can be reduced to zero via a series of gravity assists if you have infinite patience and precision, so we have three scenarios.

Assume you set it up so that that's the speed you have (375, 540, or arbitrarily close to zero) when you enter the SoI, and you have a periapsis of 30 km above Mun. How big a burn will you need to execute at periapsis to inject yourself into an orbit? The answer is 276, 359, or 184 respectively. The reason they are different is, essentially, the Oberth effect.

If you use the 375 or the 540 or the zero as the amount of your lunar orbit injection burn, that implicitly means you're expecting to be co-orbital or to orbit at the edge of the lunar SoI.

Anyway, I will have to study your math another time; work calls. One thing both our calculations ignore here is the speed at which you can produce the requisite impulse. If you have a limit of, say, TWR=2, it means your vertical climb will have far worse gravity losses.

IIRC, vertical and gravity-turn get very similar aerodynamic losses: by design, both trajectories are vertical through the vast bulk of the atmosphere.

[GoSlash27]

You're both half-wrong and half-right.

Vertical ascent = massive gravity losses. Near horizontal ascent that leads directly into a transfer burn = minimal gravity losses.

The most efficient way to do it is with a gravity turn that ends almost flat, but pops you into space at about the same time that the Mun comes over the horizon, so that you can go straight into the transfer without a circularisation burn. Maximum Oberth, minimum gravity loss.

I got invited into this discussion, and Wanderfound has already said what I would.

Vertical ascent is just plain inefficient due to gravity losses. Far more efficient to do an insertion burn to orbit and then transfer to the mun. But directly injecting into a munar transfer without stopping to circularize is the most efficient way to do it in terms of DV.

best,

-Slashy

[arkie87]

No, what I'm saying is that the Hohmann transfer's second burn of 375 m/s, or the 540 m/s you use, tell you the speed you have relative to Mun when you enter its SoI. That speed can be reduced to zero via a series of gravity assists if you have infinite patience and precision, so we have three scenarios.

Oh, I understand what you mean now. Not sure how relative velocity can be reduced to zero though via gravity assists though...

Assume you set it up so that that's the speed you have (375, 540, or arbitrarily close to zero) when you enter the SoI, and you have a periapsis of 30 km above Mun. How big a burn will you need to execute at periapsis to inject yourself into an orbit? The answer is 276, 359, or 184 respectively. The reason they are different is, essentially, the Oberth effect.

Not sure I agree with the reason they are dealing having anything to do with Oberth effect. The reason they are different is due to the fact that relative velocities are different at periapsis before the circularization burn at 30km. I havent done the math yet, but my hunch is that the 375 m/s case is moving slower at periapsis, and so, needs a smaller burn whereas the 540 m/s is moving faster (since it initially has larger relative velocity), and therefore, needs a larger burn to slow down. This effect is non-linear though, since falling faster will result in LESS deltaV than falling slow due to conservation of energy (i suppose you might consider this the Oberth effect).

If you use the 375 or the 540 or the zero as the amount of your lunar orbit injection burn, that implicitly means you're expecting to be co-orbital or to orbit at the edge of the lunar SoI.

Yes, this was the simplification i.e. rather than getting into an orbit around the Mun, i merely got co-orbital to eliminate the actual encounter's effect on things (such as gravity assist and the like).

Anyway, I will have to study your math another time; work calls.

My math should be the same as yours. Instead of combining the two equations for conservation of energy and Kepler's law of Area, I left them as is and let EES solve them simultaneously. I have since substituted one into the other and also looked the formula up (it is explicit and pretty simple, actually) so i will use that one from now on.

One thing both our calculations ignore here is the speed at which you can produce the requisite impulse. If you have a limit of, say, TWR=2, it means your vertical climb will have far worse gravity losses.

This is very true and something important to consider; however, bear in mind that this is true for the horizontal ascent as well since as you mention below, the horizontal ascent goes vertically for most of the way out of the atmosphere...

IIRC, vertical and gravity-turn get very similar aerodynamic losses: by design, both trajectories are vertical through the vast bulk of the atmosphere.

This might be true. But the main difference is that the horizontal case turns and loses vertical velocity and starts from scratch to build up velocity for Oberth whereas vertical keeps this velocity and fully utilizes oberth since it never burns away from prograde.

[arkie87]

I got invited into this discussion, and Wanderfound has already said what I would.

Vertical ascent is just plain inefficient due to gravity losses. Far more efficient to do an insertion burn to orbit and then transfer to the mun. But directly injecting into a munar transfer without stopping to circularize is the most efficient way to do it in terms of DV.

best,

-Slashy

Did you do any calculations? I think a quantitative argument holds more weight than an unproven qualitative one.

My calculations show that they are really close, despite the taboo of "vertical launches being inefficient due to gravity"... and potentially vertical ascent is better depending on the shape of the gravity turn...

[Empiro]

I think that your numbers are reasonable. When going into a very high orbit, burning straight up and then circularizing is not that different than burning sideways and then circularizing, because your final orbital velocity is relatively low, so you don't benefit too much from the Oberth effect in either case for your second burn. However, the key is that you're assuming an impulsive burn.

In the practical case, your TWR is not infinite, and every single second you're burning against gravity is 9.8 m/s delta-V lost due to gravity drag. For example, even with a very high TWR of 4, your net acceleration is 3g or ~30 m/s, and you would need to burn for over 100 seconds to accelerate to ~3300 m/s. This means that you've actually burned over 4400 m/s delta-V and you're spending an extra 1100 m/s fighting gravity. When you take into account the fact that having such a high TWR necessitates lots of extra mass in the form of engines, the balance tips even more in favor of a standard gravity turn. Even if you wind up spending a few more m/s, you've probably gained more than that delta-V from just leaving a few of your engines at home.

On the other hand, burning straight up gives you less air drag losses (because you exit the atmosphere faster). This brings things a bit more in favor of vertical burns in stock KSP, where air drag losses are relatively high, but not in FAR, where air drag accounts for a small amount of the loss in delta-V (I think it's about 200 m/s or so).

Edited December 11, 2014 by Empiro

[arkie87]

I think that your numbers are reasonable. When going into a very high orbit, burning straight up and then circularizing is not that different than burning sideways and then circularizing, because your final orbital velocity is relatively low, so you don't benefit too much from the Oberth effect in either case for your second burn. However, the key is that you're assuming an impulsive burn.

In the practical case, your TWR is not infinite, and every single second you're burning against gravity is 9.8 m/s delta-V lost due to gravity drag. For example, even with a very high TWR of 4, your net acceleration is 3g or ~30 m/s, and you would need to burn for over 100 seconds to accelerate to ~3300 m/s. This means that you've actually burned over 4400 m/s delta-V and you're spending an extra 1100 m/s fighting gravity. When you take into account the fact that having such a high TWR necessitates lots of extra mass in the form of engines, the balance tips even more in favor of a standard gravity turn. Even if you wind up spending a few more m/s, you've probably gained more than that delta-V from just leaving a few of your engines at home..

But when you burn sideways, you are losing vertical speed at the same rate until your velocity becomes large enough that centripetal lift becomes significant...similarly, when you burn vertically, you lose vertical speed until you get high enough that gravity becomes weaker, plus all of your deltaV is going prograde whereas a gravity turn part of it is wasted vertical height and part of it is horizontal...

Using the same calculations (assuming no atmospheric drag), it would take 75 seconds to get up to orbital speed for a horizontal burn during which time you will be losing about 10 m/s per second as well...

[GoSlash27]

Did you do any calculations? I think a quantitative argument holds more weight than an unproven qualitative one.

My calculations show that they are really close, despite the taboo of "vertical launches being inefficient due to gravity"... and potentially vertical ascent is better depending on the shape of the gravity turn...

Absolutely not. I have no idea how I'd approach modeling a vertical ascent to the mun.

The most definitive result would be to build a ship that has the absolute minimum DV to get you there on a prograde direct ascent, then try it with the vertical ascent.

If it runs out of steam before achieving a munar apoapsis, then you have your answer.

Even better... what is the minimum DV required to achieve a 12 Mm apoapsis?

Best,

-Slashy

Edited December 11, 2014 by GoSlash27

[arkie87]

Absolutely not. I have no idea how I'd approach modeling a vertical ascent to the mun.

The most definitive result would be to build a ship that has the absolute minimum DV to get you there on a prograde direct ascent, then try it with the vertical ascent.

If it runs out of steam before achieving a munar apoapsis, then you have your answer.

Even better... what is the minimum DV required to achieve a munar apoapsis?

Best,

-Slashy

Here are my calculations:

Assuming Kerbin has no atmosphere and is perfectly spherical so LKO is at 0 m altitude and zero rotation:

Burning vertically upward to Mun's altitude takes a total of 3348 m/s. If you include circularization burn once you get to to altitude, it takes an additional 529.4 m/s, totaling 3878 m/s.

Burning horizontally upward to Mun's altitude from surface of Kerbin takes 3352 m/s (4 m/s more!). If you include the circularization burn one you get to altitude, it takes an additional 369.8 m/s, totaling 3722 m/s.

Thus, horizontal beets vertical by 155.8 m/s, but only if we assume a transfer from 0 m altitude. Since even horizontal approach must burn vertically upward for a bit (I'd venture to say way more than 155.8 m/s), I'd say vertical ascent method is probably more fuel efficient...

Admittedly, this method requires high TWR, which is not addressed in calculations given impulse burn assumption..

[arkie87]

The most definitive result would be to build a ship that has the absolute minimum DV to get you there on a prograde direct ascent, then try it with the vertical ascent.If it runs out of steam before achieving a munar apoapsis, then you have your answer.

I plan on doing just that tonight when i get home from work, and recording the results and posting to youtube

Even better... what is the minimum DV required to achieve a 12 Mm apoapsis?

Conservation of energy: Kinetic + Potential = Kinetic + Potential: 1/2 V^2 - G*M/R = -G*M/(R+H)

Plugging in G, M, R and H, and solving for V results in 3348 m/s

[GoSlash27]

Conservation of energy: Kinetic + Potential = Kinetic + Potential: 1/2 V^2 - G*M/R = -G*M/(R+H)

Plugging in G, M, R and H, and solving for V results in 3348 m/s

Sorry, I should've been more specific. I know what the theoretical answer is for the difference between the two states. I mean the expenditure required to make it happen.

I plan on doing just that tonight when i get home from work, and recording the results and posting to youtube

I look forward to seeing the results.

Best,

-Slashy

[Empiro]

But when you burn sideways, you are losing vertical speed at the same rate until your velocity becomes large enough that centripetal lift becomes significant...similarly, when you burn vertically, you lose vertical speed until you get high enough that gravity becomes weaker, plus all of your deltaV is going prograde whereas a gravity turn part of it is wasted vertical height and part of it is horizontal...

Using the same calculations (assuming no atmospheric drag), it would take 75 seconds to get up to orbital speed for a horizontal burn during which time you will be losing about 10 m/s per second as well...

When you burn diagonally, your gravity losses are lower. For example, to experience 1g acceleration straight up, you'd need a TWR of 2. To get 1g when burning at a 45 degree angle, you'd only need a TWR of 1.41 (square-root of 2). This is because gravity has no effect on the horizontal component of your burn.

Using your numbers, to do a horizontal burn with a TWR of 4, you'd need to pitch up only about 14.5 degrees (sin(14.5 degrees) = 0.25 * 4g) to cancel out gravity. Your horizontal acceleration would be cos(14.5 degrees) = .968 * 4g = 38 m/s/s. You'd need to burn 88 seconds, or about 3462 m/s delta-V. The gravity losses are only 3462 - 3352 = 110. This doesn't even take into account that once you're at ~2200 m/s you don't need to cancel out gravity at all, since you're at orbital velocity.

[arkie87]

Sorry, I should've been more specific. I know what the theoretical answer is for the difference between the two states. I mean the expenditure required to make it happen.

Best,

-Slashy

You mean assuming non-impulse burns?

Yeah, i'm too lazy to do that... i'll just do it in KSP and record it.

[arkie87]

When you burn diagonally, your gravity losses are lower. For example, to experience 1g acceleration straight up, you'd need a TWR of 2. To get 1g when burning at a 45 degree angle, you'd only need a TWR of 1.41 (square-root of 2). This is because gravity has no effect on the horizontal component of your burn.

Using your numbers, to do a horizontal burn with a TWR of 4, you'd need to pitch up only about 14.5 degrees (sin(14.5 degrees) = 0.25 * 4g) to cancel out gravity. Your horizontal acceleration would be cos(14.5 degrees) = .968 * 4g = 38 m/s/s. You'd need to burn 88 seconds, or about 3462 m/s delta-V. The gravity losses are only 3462 - 3352 = 110. This doesn't even take into account that once you're at ~2200 m/s you don't need to cancel out gravity at all, since you're at orbital velocity.

Yes, that is true. But you also also not accounting for the amount of time it takes to climb vertically before you can begin gravity turn (which you then waste by turning sideways). And on the other end, gravity gets weaker faster if you burn vertically upward.

I wouldn't be surprised if the two approaches are close... and if they are within 250 m/s or so, i'd say vertical has the advantage due to simplicity, but obviously, to each his own.

The main point of this thread was to show that a direct mun transfer via vertical burn is easy to do and more convenient when gravity turning is hard due to lack of control, and might even be more fuel efficient (especially in career mode before you have reaction wheels and control surfaces).

I'll post a link to the youtube video once i make it

Thanks for the respectful discussion guys

Edited December 11, 2014 by arkie87

[arkie87]

Yes, that is true. But you also also not accounting for the amount of time it takes to climb vertically before you can begin gravity turn (which you then waste by turning sideways). And on the other end, gravity gets weaker faster if you burn vertically upward.

I wouldn't be surprised if the two approaches are close... and if they are within 250 m/s or so, i'd say vertical has the advantage due to simplicity, but obviously, to each his own.

The main point of this thread was to show that a direct mun transfer via vertical burn is easy to do and more convenient when gravity turning is hard due to lack of control, and might even be more fuel efficient (especially in career mode before you have reaction wheels and control surfaces).

I'll post a link to the youtube video once i make it

Thanks for the respectful discussion guys

So i made a simple program in Matlab that simulates non-impulse burn. You specify TWR, and it will calculate acceleration, velocity, and displacement. Both cases ignore atmospheric drag forces and assume TWR is constant (no mass is lost--sorry, i was lazy).

For vertical ascent, it calculates vertical burn starting from surface until it reaches velocity necessary to reach Mun altitude (3348.2 m/s).

For horizontal ascent, it calculates thrust angle necessary for thrust to counteract gravity minus centripetal acceleration (once velocity gets above orbital, it switches to zero degrees rather than go negative ) at sea level, then integrates horizontal momentum until the velocity necessary to obtain apoapsis at Mun's altitude is reached.

Both burns are timed. The expended deltaV is proportional to time, so small times mean less deltaV.

Below are the results of excess/wasted DeltaV of vertical burn relative to horizontal burn as a function of TWR

Horizontal is significantly more efficient, unless TWR is high.

So, it seems the Kerbals were right when they said to add MOAR BOOSTERS!

Edited December 11, 2014 by arkie87

[Superfluous J]

I'm using FAR (the only mod relevant to physics) and stock parts. I can send you the craft files if you really want, but they are kinda difficult to fly with a gravity turn (no torque).

I've never actually done this in FAR. The upward-thrusting rocket would benefit from being able to pack on the TWR but the sideways-thrusting rocket would benefit from being able to turn quicker. Most of the 1100dV to LKO you save in FAR is due to not having to fight gravity as much.

I would recommend making your own simple craft with about 7 km/s deltaV -- orange fuel tank + mainsail + reaction wheel + control surfaces and small fuel tank + lv909 and command pod should be close?

I did do this. That's why I'm so confident in the results. I did it well over a year ago, before I even recorded one KSP video.

If you make a video, mention me for giving you the idea! I want to be famous!!

Dozens of fans, you'll have! Dozens I say!

[arkie87]

I've never actually done this in FAR. The upward-thrusting rocket would benefit from being able to pack on the TWR but the sideways-thrusting rocket would benefit from being able to turn quicker. Most of the 1100dV to LKO you save in FAR is due to not having to fight gravity as much.

I did do this. That's why I'm so confident in the results. I did it well over a year ago, before I even recorded one KSP video.

Wait, you did it already, and which method won?

Dozens of fans, you'll have! Dozens I say!

Except that this method doesnt work, so ill have haters not fans

[Superfluous J]

Wait, you did it already, and which method won?

Why, the method I champion, of course. Getting to orbit first and then transferring to Mun.