has anyone calculated the "easy" Moho transfer?

[Laie]

For those who don't know it, the easy way of getting to Moho:

• leave when Kerbin (you know, your homeworld) is just passing it's AN/DN with Moho.
  
• make it so that you match Moho's inclination from the start, and that your solar periapsis just touches Moho's orbit.
  
• when you reach your solar periapsis, change your apoapsis (and hence your orbital period) so that you'll meet Moho the next time you come around.
  

That approach isn't exactly the fastest, you take at least one extra orbit around the sun (then again, orbits down there don't take very long). However, it is easy to do and probably the most energy-efficient approach. It has the side benefit that launch windows come up twice a year.

Now, the last time I did this I spent a lot of time playing with the maneuver until everything was just right. Especially, leaving Kerbin at just the right time proved difficult: give and take an hour or two already makes for a sizable plane change. I wrote down the data, and it must be around here.... somewhere... if only I could find it.

Has anyone taken notes and is better at notekeeping than I am, or has someone already calculated that transfer, or *cough* would someone who knows how to do it please do so and post the numbers?

[LethalDose]

There are two problems with what you're asking:

The first, and largest, is this vaguely defined nightmare step:

make it so that you match Moho's inclination from the start, and that your solar periapsis just touches Moho's orbit.

Unless you can define where, when, and how this burn is made, we can't help you calculate it.

Second, and less of an issue, is the burn at periapsis to get the Moho encounter (I refer to this as the "synchronization burn" here, but call it whatever, that's not official). this is a problem because the magnitude of the synch burn is going to vary on Moho's location, and the expenditure of energy at this burn in the Sun's SOI will be somewhat less efficient than the expenditure of energy close to Moho because Oberth. The two burns (sync and orbital capture) both basically add up to a single total, but the way you spend it is going to vary which screws up the calculations.

Also, IIRC, Moho's semi-major axis isn't perpendicular to the Moho-Kerbin AN-DN axis, so the altitude of the synch burn varies as well, which further mucks up a nice clean calculation.

So without more data, you can't calculate it.

Would some worst-case scenario calculations work for you?

Addendum: You may actually be able to avoid the first problem if you don't make a inclination correction, understanding the three following caveats:

• You'll have a pretty narrow window when you want to exit Kerbin's SOI/make your initial burn.
  
• Even if you do that right, it'll still be tricky to get the encounter right on the synch burn.
  
• you'll come in with a lot of offset vertical velocity.
  

I think a clever astro-navigator can mitigate some of those issues, but it's still a bit trick to fly.

Edited January 16, 2015 by LethalDose

[cantab]

The simplest way will be to fly it. Keep in mind that the most efficient approach will be to launch into a somewhat inclined Kerbin orbit, such that you can make a purely prograde ejection burn that puts you into a transfer orbit in the plane of Moho's.

Of course it's even more efficient to snag a few gravity assists. Eve can help of course, and even Moho itself can be used to reduce your insertion burn. Look at the real-world MESSENGER probe, which made flybys of Earth, Venus twice, and Mercury three times before entering Mercury orbit on the fourth encounter!

[Laie]

There are two problems with what you're asking:

The first, and largest, is this vaguely defined nightmare step:

Huh? I thought that was pretty well-defined: there's only two possible dates (I expect them to come around on the same day every year, just like solistice) where you can match planes during your transfer burn. The destination is a point touching Moho's orbit on the far side of the sun, presumably also the corresponding AN/DN with Kerbin.

I did not even think about what orbit to start from -- but assuming a 100km equatorial is as good as anything else.

Second, and less of an issue, is the burn at periapsis to get the Moho encounter (I refer to this as the "synchronization burn" here, but call it whatever, that's not official).

This is a non-issue even with KSP's stock toolbox: just put down a maneuver, pull retro, and watch what happens to the closest approach markers. No calculation required. If necessary, put down a second maneuver to look several orbits ahead (though IIRC it should hardly ever take more than two orbits to get your encounter).

The maneuver that I'm interested in is the one that gets me out of Kerbin. If that one is done well, everything else is a piece of cake. Problem is that I can eyeball it as much as I want, I always end up being off by one degree or more (which is expensive down there!).

[Aethon]

Here you go.

http://forum.kerbalspaceprogram.com/threads/26656-Delta-V-to-reach-Moho-orbit/page2?highlight=moho

Scroll down to Macollo's post. Mine under it has the date.

The first time this occurs ( IIRC the AN ) is ( Earth ) Year 1 Day 21, on or about Hour 17.

Edited January 16, 2015 by Aethon

[Superfluous J]

Huh? I thought that was pretty well-defined: there's only two possible dates (I expect them to come around on the same day every year, just like solistice) where you can match planes during your transfer burn. The destination is a point touching Moho's orbit on the far side of the sun, presumably also the corresponding AN/DN with Kerbin.

He's not talking about the orientation of Moho's ORBIT, he's talking about where Moho is on its orbit.

Burning to lower your Apoapsis in Sunar space is less efficient than when you're flying by Moho and taking advantage of the Oberth Effect. It's not MUCH, but it IS less efficient.

I personally don't care about it. I go to Moho using this exact method with the exception that I don't bother to match planes. I do that all when I'm coming in to Periapsis around Moho, as I'm slowing down.

EDIT: Oops I answered your second question as if it was your first. I blame how the forum won't let you quote quotes.

[LethalDose]

Huh? I thought that was pretty well-defined: there's only two possible dates (I expect them to come around on the same day every year, just like solistice) where you can match planes during your transfer burn. The destination is a point touching Moho's orbit on the far side of the sun, presumably also the corresponding AN/DN with Kerbin.

I did not even think about what orbit to start from -- but assuming a 100km equatorial is as good as anything else.

Wanting to be in the proper inclination from the start is exactly what makes this an nightmare step, but at least now you've actually clarified you wanted it from the exit burn.

This was not clearly defined when you asked "has someone calculated that transfer".

If this is what you want, then starting from an equatorial orbit is moronic. It would make way more sense to launch directly into an inclined orbit from KSC, but determining that target inclination is, once again, a freaking nightmare. You need to calculate what the target velocity in the normal direction, but that's going to change with different exit burns. The calculation gets really ugly really fast.

In fact, I think you may even want the inclination around Kerbin to be in the opposite direction as Moho's inclination, since you're ejecting out the back of Kerbin.

Either way, it's a mess.

This is a non-issue even with KSP's stock toolbox: just put down a maneuver, pull retro, and watch what happens to the closest approach markers. No calculation required. If necessary, put down a second maneuver to look several orbits ahead (though IIRC it should hardly ever take more than two orbits to get your encounter).

The maneuver that I'm interested in is the one that gets me out of Kerbin. If that one is done well, everything else is a piece of cake. Problem is that I can eyeball it as much as I want, I always end up being off by one degree or more (which is expensive down there!).

At least 5thHorseman already handled this answer for me.

He's not talking about the orientation of Moho's ORBIT, he's talking about where Moho is on its orbit.

Burning to lower your Apoapsis in Sunar space is less efficient than when you're flying by Moho and taking advantage of the Oberth Effect. It's not MUCH, but it IS less efficient.

I personally don't care about it. I go to Moho using this exact method with the exception that I don't bother to match planes. I do that all when I'm coming in to Periapsis around Moho, as I'm slowing down.

And he even addresses not making the inclination change at all as I did in the addendum.

[Taki117]

Took a lot of searching, but I found what I was looking for.

http://forum.kerbalspaceprogram.com/threads/61478-Oh-bugger-Injection-burn-at-Moho?p=835667&viewfull=1#post835667

[cantab]

If this is what you want, then starting from an equatorial orbit is moronic. It would make way more sense to launch directly into an inclined orbit from KSC, but determining that target inclination is, once again, a freaking nightmare. You need to calculate what the target velocity in the normal direction, but that's going to change with different exit burns. The calculation gets really ugly really fast.

Calculation the ideal inclination is indeed a nightmare, but working it out empirically shouldn't be too hard. Make a guess, see what the ejection burn would be, refine that guess. If you're happy to use Hyperedit or fuelhack to support your working out it won't even be all that hard.

We see from the thread Taki linked to - and excellent find btw - that the ejection inclination isn't that much. Looks like maybe 10 or 20 degrees in the right orientation will do it.

[smjjames]

Theres also whether you intend to stop at Moho or just fly by because you need a LOT of DeltaV to stop at Moho, and theres no atmosphere to help you brake.

[WuphonsReach]

Theres also whether you intend to stop at Moho or just fly by because you need a LOT of DeltaV to stop at Moho, and theres no atmosphere to help you brake.

My personal rule of thumb for stopping at Moho is 4200-4500 DV... Alexmoon's calculator puts it at only 3200-3300. But that would require perfection on the ejection burn and a perfect low velocity intercept and I never manage to pull that off.

[capi3101]

The question I never got answered as far as macollo's approach to getting to Moho is on what day it's properly aligned for a transfer. I mean, yeah it happens twice during a Kerbin year, but what's the initial day and how often is the period between windows? Better yet, how could I calculate that information for myself?

[FlyingPete]

It makes sense to me- first step is to launch yourself into a solar orbit with the same inclination as Moho, by ejecting from Kerbin orbit at the AN or DN of Kerbin/Moho. At this point, the position of Moho doesn't matter, just that the inclinations are the same, and your Sun Pe touches Moho's orbit.

From here, it's exactly the same process as rendezvousing with a spacecraft- do a prograde or retrograde burn at periapsis so that next time you come around, Moho is there to meet you. There will be a rare launch window where Moho is in the right position to meet you on the first pass, making the second part unnecessary.

[Aethon]

If you have a ship in orbit, you can set Moho as its' target, and pull out a dummy maneuver node far enough to escape Kerbin. This will allow you to see the AN/DN between your current Kerbin orbit and the plane of Mohos' orbit. Obviously it happens twice a year.

Wait for Kerbin to be very close to the AN ( Earth Year 1 Day 21, on or about Hour 17). The next time you pass the place where you think you want your exit burn to start (retrograde to Kerbins' orbit), set up your actual escape burn maneuver node so it just touches Mohos' solar periapse, and it has enough anti- normal (south) so that your inclination relative to Mohos' orbit is zero. Combining your inclination change with your escape burn saves mucho dv.

Thanks to improved chase camera, you can see System Shuttle Argos and IPEX Hydras' anti-normal component built in to the Kerbin escape burn. No calculations necessary. It's all handled by the maneuver node system. After the burn, Argos refuels Hydra, undocks, and burns retrograde, returning to Kerbin for an aerobrake, refuel and reuse. This burn was around 2300 m/sec.

After you pass your (and Mohos') solar periapse, set up a retrograde maneuver burn on the periapse ( which you just passed ) so that you will encounter Moho on your next orbit, at a dv cost of around 1800m/sec. This has the added bonus of decreasing the relative velocities in the eventual Moho encounter. My Moho capture burn cost around 1000 m/sec.

Hope this helps.

Edited January 16, 2015 by Aethon

[LethalDose]

Calculation the ideal inclination is indeed a nightmare, but working it out empirically shouldn't be too hard. Make a guess, see what the ejection burn would be, refine that guess.

To the OP, I'd say this is the best answer: empiricism.

If you decide to try to find what inclination works best for your ejection, share with us.

[Laie]

To the OP, I'd say this is the best answer: empiricism.

(sigh)

Alrighty. Is it possible to at least determine the date/time of inclination nodes with Moho? Or, even better: the position of my AN around Kerbin that would line up with Kerbin, Sun and Moho's AN once the time is right? Assuming that I start over Kerbin's equator (1), I should strive to do the burn at the precise moment when Kerbin goes past the node, amirite? That should be within my math skills, actually, but I have no idea how to get the path parameters of a planet to begin with.

(1) don't tell me wether that's smart or not. I need some data to start with, and within the precision of a burn, the equator is just as good a starting point as anything else. It should be easier to calculate, I guess.

[Superfluous J]

Assuming that Kerbin's year is actually 1 revolution around the Sun* then all you need to know is where Kerbin is radially around the Sun at a certain day, and what angle the AN and DN of Moho are. Then simple math will tell you at what date and time every year Kerbin will be at those angles.

*That's not assured. I think 1 6-hour day is not what we here would define as a day; the Sun does not rise at the same time each "day."

[Kryxal]

Here's the answer you're looking for, last line of the quote:

Here you go.

http://forum.kerbalspaceprogram.com/threads/26656-Delta-V-to-reach-Moho-orbit/page2?highlight=moho

Scroll down to Macollo's post. Mine under it has the date.

The first time this occurs ( IIRC the AN ) is ( Earth ) Year 1 Day 21, on or about Hour 17.

Just add multiples of a Kerbin year (and convert to 6-hour days if you like).

[Aethon]

Jees.. Is this thing on?? TAP, TAP, TAP! Check 2... Check 2... SSSSibilant, sibilant. PHHHHHT.... PHHHHT.

To determine when Kerbin crosses Mohos' AN/DN:

If you have a ship in orbit, you can set Moho as its' target, and pull out a dummy maneuver node far enough to escape Kerbin. This will allow you to see the AN/DN between your current Kerbin orbit and the plane of Mohos' orbit. Obviously it happens twice a year.

The date this happens the first time:

The first time this occurs ( IIRC the AN ) is ( Earth ) Year 1 Day 21, on or about Hour 17.

If I'm still not making myself clear, you can't contact me directly (PM) and I can explain further.

Edited January 18, 2015 by Aethon

[PLAD]

Here's how you can calculate when Kerbin will pass through the plane of any other planet's orbit, using Moho as an example. I always use Earth time, 365/24.

In the official KSP wiki you can find the orbital parameters for all the KSP bodies. For Moho we find that its Longitude of Ascending Node (LAN) is 70 degrees. (We don't know what direction 0 degrees is yet but we'll get to that.) Moho's orbit crosses the plane of Kerbin's orbit twice, once going up and once going down. These two must be exactly 180 degrees apart, so we now deduce that Moho's descending node is at 250 degrees. This means if you leave Kerbin when it is at 70 degrees or 250 degrees you will be exactly in the plane of Moho's orbit. We determine these times by noting that the wiki says Kerbin's Mean Anomoly at 0s UT is 3.14 radians. This equals about 179.9 degrees, so now we know where Kerbin starts. Since Kerbin's orbit is circular it moves the same speed at all times, which is 360 degrees every 9,203,545 seconds, or 106.52251 Earth days. This is 3.3795673 degrees per day. Therefore Kerbin first crosses Moho's descending node at (250-179.9)/3.3795673 = 20.7423 days after game start and crosses Moho's ascending node at (360+70-179.9)/3.3795673= 74.003556 days after game start. Since the game starts at day 1 (and not day 0!) that means you can first launch from Kerbin into the plane of Moho's orbit on day 21.7423 and day 75.0036, and every 106.5225 days after those two dates.

The reverse question is trickier, when is Moho in the plane of Kerbin's orbit? We know it will be when Moho is at 70 or 250 degrees, but because Moho's orbit is not circular it does not move the same distance every day. You have to use a Kepler algorithm to get the exact times. Moho first crosses Kerbin's orbital plane on days 13.128 and 25.09, and every 25.6453 days after those.

An even more interesting challenge is determining when, for instance, Eve crosses the plane of Moho's orbit. Now you have to cross-multiply the normal vectors of the two orbit planes (which you find by cross-multiplying the radius and velocity vectors at a chosen time) to get the angle of the nodes, then use a Kepler algorithm to find when the planets are first at those points. I leave that rather hairy problem as an exercise for the reader.

Note that none of these tell if Moho will actually be there when you arrive at its orbit. The best way to determine when to leave Kerbin for Moho is using a pork-chop plotter like Alex Mun's.

[Laie]

@Aethon: See 5thHorseman, above.

Eyeballing is nice and fine, but I want to at least calculate the precise date. Guess that will also help me to figure out whether a year is really one orbit of Kerbin.

Wiki says that one orbit of Kerbin around the Sun lasts 9203545 seconds -- that this number doesn't end in several zeroes already tells me that "one calendar year" can't be "one orbit". Calculator says one orbit takes 426.09 calendar days. That's a good thirty minutes extra per year. This is going to be fun. Probably one has to calculate in seconds since epoch, then convert to a proper date.

BTW, one rotation of Kerbin lasts exactly one calender day; together with Kerbin's movement around the Sun, that means that each day, the sun will rise ~51 seconds later than on the previous day; every year of 426 calender days has only 425 sunsets.

[LethalDose]

... but I have no idea how to get the path parameters of a planet to begin with.

Have you not been to the wiki? The planetary entries have every orbital parameter you could want. There's a link at the top of literally every forum page...

(1) don't tell me wether that's smart or not. I need some data to start with, and within the precision of a burn, the equator is just as good a starting point as anything else. It should be easier to calculate, I guess.

The emphasized part is factually incorrect. You can spend < 200 m/s of dV to launch into the correct orbital inclination off the pad, or you can spend nearly 1000 m/s (possibly more) to adjust your orbital inclination during your escape burn.

I won't tell you if that's smart or not.

[Laie]

Here's how you can calculate when Kerbin will pass through the plane of any other planet's orbit, using Moho as an example. I always use Earth time, 365/24.

Thanks a lot. I'll have to do that gain in Kerbin days, but given your directions this will be easy.

And yes, there is no day zero... So 426 calender days in a year are only really 425 days? That's besides the point for the topic at hand, but I'm trying to bend my mind around the concept as a whole.

Note that none of these tell if Moho will actually be there when you arrive at its orbit. The best way to determine when to leave Kerbin for Moho is using a pork-chop plotter like Alex Mun's.

For this transfer, this doesn't matter. You just start slowing down when you arrive at that point regardless of whether Moho is there or not. If you do it just right, Moho will be there the next time you come around. The game's encounter indicators make that part easy. Rationale: Moho has such a tiny gravity that it barely helps you with the capture; doing a good deal of the deceleration(?) burn in it's absence will not cost much delta-V.

- - - Updated - - -

Have you not been to the wiki?

The emphasized part is factually incorrect. You can spend < 200 m/s of dV to launch into the correct orbital inclination off the pad, or you can spend nearly 1000 m/s (possibly more) to adjust your orbital inclination during your escape burn.

a) Where do the figures in the wiki come from?

whether I launch into the right plane, or raise apoapsis and then change planes, or do nothing of the sort -- that's not important (1). I need to fly off at a certain time in a certain direction and with a certain speed; that's what I'm interested in (and PLAD has helped me big time with the first one). How I make my ship actually fly off is a solved problem.

(1)Besides, even a pre-inclined orbit will probably have the maneuver node somewhere near the equator.

[PLAD]

Edited January 18, 2015 by PLAD
argh, didn't attach the quote properly

[LethalDose]

a) Where do the figures in the wiki come from?

whether I launch into the right plane, or raise apoapsis and then change planes, or do nothing of the sort -- that's not important (1). I need to fly off at a certain time in a certain direction and with a certain speed; that's what I'm interested in (and PLAD has helped me big time with the first one). How I make my ship actually fly off is a solved problem.

(1)Besides, even a pre-inclined orbit will probably have the maneuver node somewhere near the equator.

a) I believe they come straight from the code. I've never once found one inaccurate value there.

Your OP stated you wanted the plane change done on the initial burn. If you're changing your mind about that now, whatever.

(1) Unlikely if you're still planning on aligning your Kerbin orbital AN with the Kerbin-Moho AN, especially if you correct for the required burn time.

For this transfer, this doesn't matter. You just start slowing down when you arrive at that point regardless of whether Moho is there or not.