When is the annual Moho transfer, again?

[Laie 1,408]

The easy, reliable and repeatable method of getting to Moho:

1. leave Kerbin when it is at it's AN with Moho.
2. set up a transfer orbit around the Sun so that you vessel's solar PE touches Moho's orbit
   Important: fix inclination already at departure from Kerbin!
3. at solar PE, burn retrograde so you encounter Moho the next time you come around.

That's reasonably fast, you usually get to Moho within less than a Kerbin year. More often than not, it's even economical: Proper transfer windows that require less dV only crop up once every couple of years.

However, I keep forgetting the date. I've written it down somewhere, lost the notes, can't find an old forum thread, and every time I want to go to Moho I have to figure it out almost from scratch.

Q: is there a method to quickly determine the date using in-game tools? Quicker than lining things up in map view and fast-forwarding until Kerbin is in the right place?

(BTW, using the above method I get a departure time late on day 82.)

[Streetwind 4,947]

25 minutes ago, Laie said:

Q: is there a method to quickly determine the date using in-game tools?

Unless you install a mod, no. Transfer windows are too complex to eyeball the dV cost without doing the math.

[Superfluous J 15,266]

26 minutes ago, Laie said:

Q: is there a method to quickly determine the date using in-game tools? Quicker than lining things up in map view and fast-forwarding until Kerbin is in the right place?

Other than lining up the orbits so you don't need to fast forward... No. I don't know of one. I mean, you need to fast forward to get there but you'd need to do that anyway.

[Laie 1,408]

37 minutes ago, Streetwind said:

Transfer windows are too complex to eyeball

In this particular case, eyeball precision goes a long way. I'm only looking for Kerbin's AN with Moho, after all.

[Zhetaan 922]

On 8/2/2020 at 5:45 AM, Laie said:

In this particular case, eyeball precision goes a long way. I'm only looking for Kerbin's AN with Moho, after all.

Since you got a departure date of day 82, wouldn't the subsequent windows for this method be multiples of 426 days (one Kerbin year) after this date?  You're not transferring directly to Moho, but rather a point on Moho's orbit, and since that is a fixed orbit without perturbation, there's no synodic period:  Kerbin will pass its ideal alignment with that point once every year.

Edited August 3, 2020 by Zhetaan

[Laie 1,408]

49 minutes ago, Zhetaan said:

You're not transferring directly to Moho, but rather a point on Moho's orbit, and since that is a fixed orbit without perturbation, there's no synodic period:

Yes, ony any given Kerbin year, day 82 is Leave for Moho Day. Or perhaps Day 83, zero hours.

[Grogs 485]

I *think* the way the KSP geometry is laid out the value you want is the longitude of the ascending node (LAN). I'm assuming Kerbin is at longitude 0, however that's calculated, at the beginning of the year. According to the Wiki, Moho has an LAN of 70 degrees. 70 / 360 * 426 days = 82.833 days, so your eyeballing was pretty good if that's right.

 

BTW, since you're not targeting Moho the DN should work too. The DN would be at 250 degrees, or 295.833 days into they year.

[Zhetaan 922]

1 hour ago, Laie said:

Yes, ony any given Kerbin year, day 82 is Leave for Moho Day. Or perhaps Day 83, zero hours.

Oh!  I misunderstood earlier; you know that it's Day 82, but in case you forget again, you need a way to determine that Day 82 is the day that you want without needing to memorise it or rely on pre-calculation.  In that case, I can think of two answers.  The first is to use manoeuvre nodes to plot an orbit that just barely leaves Kerbin's sphere of influence, set up a transfer to Moho's ascending node, and note the time-to-node.  That is subject to inaccuracies because the thing in solar orbit is necessarily not precisely in Kerbin's orbital track, but it's probably close enough.  The only way I can think of to get more accurate using in-game tools is to plot an escape burn to Moho's ascending node altitude and start pressing the + Orbit button until you get a burn that actually goes there.  You'd need to tweak the node location in order to keep your escape pointed Kerbin-retrograde and you'll likely get joint fatigue in your knuckles from all of the clicking, but it should work.

This depends on having something in orbit that you can use to plot these manoeuvres, whether it's your Moho vehicle or a random low-orbit satellite, but reference rockets that you put in orbit are arguably in-game tools, so there it is.

Granted, you asked for fast, not accurate, but aside from setting this up in advance, I can't think of a faster way using only the in-game tools.

31 minutes ago, Grogs said:

I *think* the way the KSP geometry is laid out the value you want is the longitude of the ascending node (LAN). I'm assuming Kerbin is at longitude 0, however that's calculated, at the beginning of the year. According to the Wiki, Moho has an LAN of 70 degrees. 70 / 360 * 426 days = 82.833 days, so your eyeballing was pretty good if that's right.

This is a situation where you're getting a good answer for the wrong reasons.

In fact, the only reason that this works is because Kerbin is in a zero-eccentricity, zero-longitude-of-ascending-node, zero-argument-of-periapsis orbit that can thus directly relate the true anomaly value of Moho's longitude of ascending node to the mean anomaly value of its own orbital position without any correction at all.

However, Kerbin starts at π radians at epoch, not zero.  The reason it works anyway is because it sets up a transfer orbit:  you want to start on the opposite side from the destination, so adding π radians is a necessary step.  If Kerbin had had any other starting location, then this would not have worked.

Edited August 3, 2020 by Zhetaan

[Laie 1,408]

36 minutes ago, Grogs said:

According to the Wiki, Moho has an LAN of 70 degrees. 70 / 360 * 426 days = 82.833 days, so your eyeballing was pretty good if that's right.

because of what @Zhetaan point out in the next post, it never occurred to me to try that simple calculation.

38 minutes ago, Grogs said:

BTW, since you're not targeting Moho the DN should work too. The DN would be at 250 degrees, or 295.833 days into they year.

In principle yes, in practice it's more costly: starting at AN, you can go to Moho's PE (well, close enough) while when starting at DN, you'll go to it's AP. I haven't tried, but expect it to make 500-1000m/s difference.

17 minutes ago, Zhetaan said:

Oh!  I misunderstood earlier; you know that it's Day 82, but in case you forget again,

When I first came here, I had forgotten... again. I just hoped that someone would provide the right answer like, real quick.

Then again, figuring it out didn't take all that long. This time around I took a few screenshots and put up a tutorial page on the wiki. Trouble with getting to Moho is kind of an FAQ around here, after all, I guess a prefab tutorial may come in handy.