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LKO/LEO/LGO inclination effect on interplanetary transfers


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Hello guys,

I have an "advanced" question.

After playing different solar systems for a while (mainly Galileo planet pack and RSS) I'd like to optimize my interplanetary transfers. On both Earth and Gael, the space centres are not exactly on equator, which leads to non-zero-inclination low orbit right after launch. In RSS, I am capable of launching pretty much exactly into the plane of the Moon, or any space station I placed in Earth orbit, this is not where my problem lies.

The problem is that I am not sure which is the ideal orbit for interplanetary transfers. Even if the orbital planes of the planets are the same (or pretty similar), I have no idea when to launch for the optimal transfer. I have KER installed for all the information it provides, but I have no idea which info to look for in this case.

Thanks for any insight,

Michal.don

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49 minutes ago, michal.don said:

The problem is that I am not sure which is the ideal orbit for interplanetary transfers. Even if the orbital planes of the planets are the same (or pretty similar), I have no idea when to launch for the optimal transfer. I have KER installed for all the information it provides, but I have no idea which info to look for in this case.

This is worded confusingly, so allow me to rephrase it in my own words, just so we can be sure we are talking about the same thing:

You are wondering which orbit around Gael is optimal for a transfer to another planet. If you knew this orbit, you would be able to, by yourself, figure out when to launch in order to achieve this orbit without a plane change.

 

The most important thing to understand here is that inclination around Gael does not properly translate into inclination around the Sun. Even if you leave Gael in a heavily inclined orbit - if you exit Gael's SOI directly at the nadir point, for instance - you would still only have a small few degrees of inclination around the Sun. You'd be at most the radius of Gael's SOI in distance away from the equatorial plane of the Sun. To make matters worse, if you wanted to do any math to that end, you would need to do the math for the specific position of Gael in its orbit around the Sun, because the result would only be valid for that specific position. As Gael travels, the result changes constantly. And then there's the fact that if you are in an inclined orbit, you are no longer burning perfectly prograde or retrograde to Gael's solar orbit for your ejection burn, and you may potentially lose more efficiency through this than you gain from making the plane change smaller.

This has two immediate results. One, trying to leave in an inclined Gael orbit very rarely ever has the effect you want on your Sun orbit; and two, computation of the optimal escape burn gets arbitrarily complex when allowing for inclined orbits.

As a result of this, interplanetary transfer planning tools like Alexmoon's web calculator (and its ingame mod equivalent, Transfer Window Planner, which by the way does support modded solar systems) usually make the assumption that you are departing from a zero-inclination equatorial orbit. (I say "usually" because I make no claim to know all such tools and, who knows, there may or may not be one that I have personally never used that does compute inclined departure orbits.) The difference in dV cost is generally negligible.

In this case I would strongly suggest the KISS principle. Launch into a zero inclination orbit, and then use a tool to calculate the optimal transfer from that starting point. It works reliably, without much effort, and has no significant downsides.

Edited by Streetwind
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1 hour ago, Streetwind said:

This is worded confusingly, so allow me to rephrase it in my own words, just so we can be sure we are talking about the same thing:

You are wondering which orbit around Gael is optimal for a transfer to another planet. If you knew this orbit, you would be able to, by yourself, figure out when to launch in order to achieve this orbit without a plane change.

Yeah, sorry about that, I'm a bit sick now and my english skills are probably not at their best......

It's not about the inclination value of the orbit, but rather about the AN/DN position. From the KSC, I can reach an orbit with minimal inclination of cca 8°, without a plane change burn. The question is - are certain 8° inclined orbits better for interplanetary transfers than other 8° inclined orbits?

I am using the Transfer Window Planner for mission planning, and the results don't vary much from actual delta-V expended - about 100 m/s for close transfers and 200 m/s for longer ones (in a 3.2x rescaled system - so about 5-10%). Sometimes a plane change burn is required in solar orbit, but that would be neccessary if I came from equatorial orbit, as well.

To sum up, I'm wondering:

 

1) Is it better to change the orbital plane to an equatorial orbit before the transfer?

2) Is there a way to recognise which AN/DN position is the best for a given transfer? Then I can plan the launch time accordingly and reach the correct orbit.

3) Is this all hassle worth it, or should I just forget it and pack the extra ~200 m/s worth of fuel, just in case I end up in a sub-optimal orbit?

 

I hope I manged to explain myself a little bit better, and not the complete opposite :)

Thanks,

Michal.don

 

 

P.S. When I was playing RSS/RO, I had a similar problem. I "solved" it by launching all my interplanetary stuff into an orbital plane of the Moon, which kind of worked, but for some transfers, it cost several hundred m/s more than the "optimal" transfer. The problem I'm thinking about now is the same, just not that costly in terms of dV.

Edited by michal.don
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1 hour ago, michal.don said:

1) Is it better to change the orbital plane to an equatorial orbit before the transfer?

2) Is there a way to recognise which AN/DN position is the best for a given transfer? Then I can plan the launch time accordingly and reach the correct orbit.

3) Is this all hassle worth it, or should I just forget it and pack the extra ~200 m/s worth of fuel, just in case I end up in a sub-optimal orbit?

1.) By making a plane change in low orbit? No, that's too expensive. By launching directly into an equatorial orbit (i.e. combining the plane change with the launch burn for big savings)? ...Maybe. It definitely makes everything a whole lot easier. But "better" in the sense of "costs less fuel"? I have no clue how to prove one or the other. Maybe @OhioBob does.

2.) Gut feeling says: with the nodes at 90 degrees from your ejection angle (the point where you put the maneuver node). This maximizes your elevation above or below the solar equatorial plane, for what tiny little actual effect that has. Then, under the assumption that it actually is beneficial to leave from an inclined orbit (which I can neither prove nor disprove), it maximizes that benefit.

3.) I already told you my opinion: keep it simple and launch into an equatorial orbit in the first place. :P For me it would definitely not be worth the hassle.

Edited by Streetwind
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It depends where you‘re going. Remember that inclination changes take less delta v at apoapsis? This also applies to interplanetary transfers (if they‘re off plane). 

The big difference: Is your destination closer or farther from the sun? If it is closer, you should try to correct as much inclination as possible with your Gael-exit. If it‘s farther from the sun, you don‘t need to worry about your initial inclination too much, as the correction of inclination can be done cheaply on the way or on arrival.

Here‘s a trick to launch right into the perfect inclination: Have a dummy craft in a low equatorial orbit and use it to plot a transfer node. Now launch your craft into an orbit with the same inclination and Longitude of ascending node as the plotted trajectory of the dummy craft.

Edited by Physics Student
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What I've generally done is:

I'll place a reference satellite with an MJ core in an equatorial Gael orbit. When a transfer window approaches, I'll create a transfer maneuver, and look to see whether it ejects north or south of the plane. If it ejects north, I'll launch from 270 degrees before the transfer node, if south, I'll launch from 90 degrees before, such that the normal component of my parking orbit is in the same direction as the normal component of the transfer orbit.

I have no idea if this is optimal, but it's a simple rule to deal with non-equatorial launches.

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I'd use the planner-sat approach too, but would probably go a bit further, and on launch try to match the plane of the plotted maneuver's ejection from whichever planet.  If you can't match the plane because of low inclination, just launch 90 or 270 degrees before the node so as to at least match the node's position.

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I haven't seriously studied this issue, so I can't say with certainty what is optimal.  However, a typical interplanetary transfer orbit looks something like this:

fig5-06.gif

So I think it would be best to launch into a parking orbit that is inclined in the same direction as your eventual interplanetary transfer orbit.  That means the line of nodes should point toward the Sun (or at least as close as you can make it).  To achieve this you want to launch into the parking orbit when KSC is on either the prograde or retrograde side of the planet, depending on the inclination of the transfer orbit and geographic location of KSC.  For the orbit transfer shown in the figure above, and assuming KSC is located in the northern hemisphere, then you want to launch when KSC is on the prograde side.

That's my 2¢ anyway.  I make no guarantees that there isn't a better way of doing it.
 

Edited by OhioBob
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@Streetwind, @Physics Student, @OhioBob, @Starman4308, @Kryxal,

Thank you all very much for your time and answers. I think the transfer-planner-sat will be the way to go, as it is quite simple, and allows to plan a precise departure orbit. Also, the diagram is really helpful.

Hopefully it works well and it won't be neccessary to ask more questions :)

Thanks again,

Michal.don

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