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Landing on Minmus, Eve or Duna is easier than on Mun?


PatPL
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Ok, so:

Landing on Mun:

- 4400 deltaV for LKO

- 850 deltaV for Mun Intercept

- 750 deltaV for Mun Landing

- TOTAL: 6000 deltaV (my record is about 5550 deltaV with SSTM plane)

Landing on Minmus:

- 4400 for LKO

- 900 for Intercept

- 280 for landing

- TOTAL: 5580!

Landing on Duna:

- 4400 LKO

- 950 Escape velocity

- 140 Perfect collision course with Duna

- 0 Landing on chutes

- Total: 5490!!

Landing on Eve:

- 4400 LKO

- 950 Esc.Vel.

- 100 Collision course

- 0 landing

- Total: 5450!!!

In theory we should first fly to Eve, then to Duna, and then Kerbin moons (if You want to land)

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There isn't a Hohmann transfer window to Duna until about 50-60 Earth days after a new Game is started. There isn't an Eve transfer window until about 80 Earth days after that. The question becomes, "Are you willing to wait nearly 150 days for your first extraplanetary landing?"

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...

In theory we should first fly to Eve, then to Duna, and then Kerbin moons (if You want to land)

Of course, you will then have to cope with the fact, that Jeb, Bill and Bop will be stranded on Eve forever :D

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Hum, TBH I find Laythe easier than Duna , in spite of needing more dV ... mainly because the Jool SoI is huge and easier to hit than Duna one and a Laythe encounter is trivial to get in you get to Jool SoI near enough of it's equator plane ( landing on land is another issue ... ). But yeah, Duna is relatively easy ...

But anyway, no interplanetary target is easier than going to the moons of Kerbin. You get a opportunity per orbit to burn to the moon you choose ( that , if in LKO is pretty much once in 30-40 game min. ) and you can do it by eyeball ( how do you think people got to the Mun and Minmus before manouver nodes ? :D )

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Fuel-budget-wise, any body with an atmosphere is potentially much 'cheaper' to reach since you can aerocapture and aerobrake (though it usually makes up for that by being harder to reach orbit once landed; see Eve for an extreme example). But in terms of sheer simplicity of mission profile, airless bodies are ideal since you don't have to worry about that pesky atmosphere mucking things up. It will always take the same amount of effort to get to and from the surface of an airless body. It also helps that most of the airless bodies in KSP are also gravity-deficient (except Tylo, which is just nasty for dV budgets).

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Fuel-budget-wise, any body with an atmosphere is potentially much 'cheaper' to reach since you can aerocapture and aerobrake (though it usually makes up for that by being harder to reach orbit once landed; see Eve for an extreme example). But in terms of sheer simplicity of mission profile, airless bodies are ideal since you don't have to worry about that pesky atmosphere mucking things up. It will always take the same amount of effort to get to and from the surface of an airless body. It also helps that most of the airless bodies in KSP are also gravity-deficient (except Tylo, which is just nasty for dV budgets).

But then you have to deal with canceling out the horizontal velocity. On a body with atmosphere that is all done for you, so that you only have to focus on touching it down gently. I have seen alot of newbies have big issues with this, so I would say something like Duna would be easier to land on and on top of that you also have parachutes. But getting the actual encounter is definitely easiest with the mun.

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I found Duna the hardest to aerobrake/aerocapture with, especially on landing missions because of the thin atmosphere

Its all too easy to be still going way too fast when the chutes open ripping your ship into many pieces and scattering them over the landscape

As for Eve..... whats the dV like for getting your crew back?

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Hum, TBH I find Laythe easier than Duna , in spite of needing more dV ... mainly because the Jool SoI is huge and easier to hit than Duna one and a Laythe encounter is trivial to get in you get to Jool SoI near enough of it's equator plane ( landing on land is another issue ... ). But yeah, Duna is relatively easy ...

But anyway, no interplanetary target is easier than going to the moons of Kerbin. You get a opportunity per orbit to burn to the moon you choose ( that , if in LKO is pretty much once in 30-40 game min. ) and you can do it by eyeball ( how do you think people got to the Mun and Minmus before manouver nodes ? :D )

I started in .15, way before maneuver nodes, and I still think they overcomplicate the process of getting to the Mun.

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Hum, TBH I find Laythe easier than Duna , in spite of needing more dV ... mainly because the Jool SoI is huge and easier to hit than Duna one and a Laythe encounter is trivial to get in you get to Jool SoI near enough of it's equator plane ( landing on land is another issue ... ). But yeah, Duna is relatively easy ...

But anyway, no interplanetary target is easier than going to the moons of Kerbin. You get a opportunity per orbit to burn to the moon you choose ( that , if in LKO is pretty much once in 30-40 game min. ) and you can do it by eyeball ( how do you think people got to the Mun and Minmus before manouver nodes ? :D )

Duna has multiple issues, I find it a bit hard to land on as parachutes don't slow you down much but still useful.

Ike also like to mess up for you.

Minmus is nice to land on, speed is relaxing low without being annoying like Gilly.

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Wouldn't that make Gilly even easier to land (controlled crash) on since it requires very little in terms of dV? You could use Eve's areobraking to save a ton of dv, and possibly get an encounter with Gilly if done at the right time.

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Duna has multiple issues, I find it a bit hard to land on as parachutes don't slow you down much but still useful.

Ike also like to mess up for you.

Minmus is nice to land on, speed is relaxing low without being annoying like Gilly.

Depends on where you land. If you land in a low altitude region then you can pretty much land everything with enough parachutes and the right aerobraking manouver. But after they changed duna there is sadly very few of these low altitude regions anymore.

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I thought the first eve window opened on day 1. my eve misson launched on day 2 and only needed 1.5 km dv for transfer.

By Alexmoon's calculator, even if you left on Day 1 at time 0:0:0.0, you'd still need a minimum 1.38 km/s to transfer from a 100 km circular orbit around Kerbin to an intercept with Eve. The optimum transfer for /that/ window happens sometime before UT = 0.0.

If you want the kind of delta-v the Original Poster mentions, you'll either have to edit your persistence file into a negative Universal Time Value, or wait 149 Earth Days from UT = 0.0, where the plot gives an optimum 1.049 m/s for the transfer.

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The old tactic of burning for the navball horizon as soon as the Mun appears over the Kerbin horizon still applies.

It applies to all the moons in KSP, as long as you are deep in the parent body gravity well and are in the same plane of the moon or close enough. That is a consequence of the laws of physics , BTW :D ( I wonder where people got the idea that it only worked with the Mun, but I already been told that twice this week ... )

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I found Duna the hardest to aerobrake/aerocapture with, especially on landing missions because of the thin atmosphere

Its all too easy to be still going way too fast when the chutes open ripping your ship into many pieces and scattering them over the landscape

Yeah, it did that to me.

Once you've got there, landing on Eve is about as easy as landing on Kerbin. Getting to other planets, though, is harder than getting to the Mun or Minmus. And I don't mean in terms of delta-V, but rather setting up the transfer burn is much more complex.

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Care to explain why? I don't understand how it works.

Semimajor axis determines your orbital period. When you're transfering from a very low circular orbit, to a very high circular orbit, the semimajor axis of your transfer orbit approaches half that of the high orbit.

As a result, for a transfer orbit going from very small circular orbit to a very large circular orbit., you can use Kepler's third law to work out that the object in the high orbit will move about 63° along its orbit in the time it takes you to go from low to high orbit.

At the typical parking orbit over Kerbin (~100km), the angle between the center of Kerbin, and the edge of Kerbin, from your spacecraft's viewpoint, is about 60°. Which is why Munrise Burn (and minmusrise burn) both work. With the Mun (and with Duna), you're also dealing with a fairly large SOI for the target satellite, relative to the distance you're going to travel. It's not going to work as well with Eve (because Gilly's in a pretty elliptical orbit). With Jool, you kind of have to ask yourself "Why did you circularize down that low in the first place?", but most of Jool's satellites are in near-circular orbits.

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Yeah, for some time people had been noting that a Mun trip is somewhat more expensive than one to Minmus. Although I wouldn't expect a trip to Duna be easier.

Of course, you will then have to cope with the fact, that Jeb, Bill and Bop will be stranded on Eve forever :D

Bop, what are you doing with Bill and Jeb? Jool will be worried!

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Thanks maltesh, I was going to post the exact same thing ... I was just editing it out. What a heck, might as well post it anyway :

We know from the Kepler laws ( specifically the third ) that the square of the orbital period is directly proportional ( in other words T2 = C *a3 , with being the orbital period, C a constant that depends of the central body in question and a the semimajor axis of a eliptical orbit ( if circular a=r, being r the radius of the orbit ) ). This can actually be deducted from the Newton law of gravity and assuming that the central body mass is much bigger than the orbiter one, but I'll skip that part .

Now , let me put a Hoffman transfer schematic involuntarily ceded by wikipedia:

400px-Hohmann_transfer_orbit.svg.png

Let's assume, for simplicity sake, that the moon we want to reach has a circular orbit , and that has the radius of R' ( in other words the orbit in red above ). The period of that orbit will be, according with the Third Kepler law:

1) (TMoon)2 = C*(R')3

Using the same reasoning, the orbital period of the ship we want to send to the mun during the Hoffman transfer ( the yellow orbit in the picture ) , will be:

2) (TShip)2 = C*(a)3

As the semimajor axis of that orbit will be (R'+R)/2 ( half of the bigger axis of the ellipse ), 2) will become

3) (TShip)2 = C*((R'+R)/2)3

Now we assume that the starting orbit from where we start our burn ( the green one ) is deep inside the gravity well compared with the moon. This means that R'>>R and that R'+R -> R'. In other words, in this situation a = R' /2 and 3 ) becomes

4) (TShip)2 = C*(R'/2)3 = C*(R')3 /23

Combining 1) and 4 ) we get

5) (TShip)2 = (TMoon)2 /23

And

6) (TShip)2 / (TMoon)2 = 1/23 <=> TShip / TMoon =1/√(23) = 1/2√2 =√2 /4

As a Hoffman transfer only uses half of the orbit, the time it will take to go from the lower orbit to the moon one ( we will call it t ) will be TShip /2 . So

7) t *2 / TMoon =√2 /4 <=> TMoon = 4√2 t ( Yes, I skipped some steps here. Exercise to the reader )

If we call the angle to the center of the orbit between the initial and final position of the moon between the Hoffman transfer α ( and as we are assuming that the moon has a circular ( and , due to Kepler laws applied to a circular orbit, with uniform velocity ), then

8) α = 360º /4√2 ~ 64º

In other words, for being successful, the Hoffman transfer to a moon should be started when you have the target moon 26º ( 90 - 64, a angle that we will call from now on β) below the prograde marker ( Note , this assumes the moon as a singular point. In reality this 26º apply to the geometrical center of the visible face of the moon as seen from the ship )

Now, it can be proved that the angle that the prograde marker in a orbit and the planetary horizon is related with the height of that orbit above the planetary surface by the following relation:

9) h = R * ( 1/cosβ - 1 ) ( really, this is basic trigonometry. Don't force me to draw a diagram of this please :P )

with h being the height of the orbit from the surface and R the radius of the planet.

For β = 26º , cosβ ~0.896 and

10) h = R *0.116

For Kerbin this will give a ideal orbital height of around 69 km ( 600*0.116 ) where the center of the face of the moon as seen from the ship coincides precisely with the Kerbin horizon as seen from that ship , but orbits of 100 or even 120 km will be still close enough for giving similar enough results. But in general this result applies to all the moons in KSP that have low eccentricity orbits ( that excludes Gilly ): you can do a Hoffman transfer to a moon if you burn when the moon rises above the horizon of the parent body , if you are in a low enough orbit and in the same orbital plane than the target moon.

Edited by r_rolo1
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I thought the first eve window opened on day 1. my eve misson launched on day 2 and only needed 1.5 km dv for transfer.

At the start of a new save, you're about 25 days (Earth time) past the optimum midpoint of a rapidly-closing Eve transfer window. If you rush the tech tree you can probably get something headed out that way without needing an unreasonable dV budget. The next optimum comes about 140-150 Earth days into the save, when you can get there for about 1000 dV less than a day-1 Eve launch.

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