Best altitude for a Minmus Refueling station?

[Rakaydos]

I'm torn as to what altitude to put my minmus orbital refueling station (topped off by minmus Kethane) at.

If I put it at the very edge of Minmus's SoI, it takes VERY little delta V to go interplanetary. However, I'm assuming a drop to a low kerbin slingshot (front for inner system, behind for outer system) which requires canceling Minmus's orbital velocity.

With a high eastward orbit, breaking free of minmus SoI as far from kerbin as possible gives more speed as you go past Kerbin, improving the Kerbin Oberth effect. However, it takes more fuel to lift the fuel that high ver minmus, and the initial break from Minmus orbit has the weakest oberth possible in that SoI.

A lower minmus orbit increases orbital velocity, improving the launchinng ship's oberth effect from minmus. It also makes it easier to reach, both for incoming ships and for kethane shuttles. Too low, however, and heavy intersteller ships will lose fuel both getting into orbit and breaking out of it.

Thoughts? Is the high orbit best? What about a Lagrange orbit? (matching minmus's semimajor orbit and sitting just outside it' SoI)

[Yasmy]

The difference in delta-v between low Minmus orbit and near Minmus SOI is so small that I would just put the station near Minmus, where the time between encounter/departure windows is much smaller, and time from the surface to the station and back is much smaller. Pick a minimum time warp you can deal with, and place the station just above the altitude cutoff for that time warp.

50× above 6 000 m

100× above 12 000 m

1 000× above 24 000 m

10 000× above 48 000 m

100 000× above 60 000 m

[Yasmy]

Note that if you are going from Minmus to Eve or Duna, it is cheapest to go directly interplanetary, without sling-shotting past Kerbin.

If you are going to any other planet from Minmus, it is cheaper to drop into a low Kerbin slingshot.

Sorry. The above statements are correct for the Mun, not for Minmus. See my post below.

I've been meaning to write up my tutorial on when to slingshot around the parent planet vs. when to go direct

vs. when to put the refueling station around the parent. Seems to come up a lot.

(I like to know the delta-V and other practical factors, but my answer is put the refueling station where you will have the most fun with it.)

Edited February 20, 2014 by Yasmy

[Greenspan]

If your mining from Minmus I want to say where you have the station is irrelevant. You still have to get the same mass of fuel from minmus to the station to your destination. If the station is low the minmus-to-station will be low delta-v, but station-to-destination will be high, and vice versa. The only difference would be in the vehicle you use to transport from surface-to-station vs station-to-destination, if one is more efficient than then other you would want to spend more delta-v with that craft. I might be oversimplifying your question.

Also, keep in mind the orbital period will be shorter the lower you are so if you have a very low TWR interplanetary vehicle you may end up having to burn over multiple orbits, bleh!

How much Delta-V do you really save by slingshoting by Kerbin? Seems like the small amount of delta-v saved is offset by the more complected/time consuming maneuver, especially in a world were parts/fuel is free and time is the only real cost.

[Superfluous J]

I put my minmus stations at 50k height. It's far enough away for 10k time warp (as was mentioned already) and is also far enough away so that as long as you dock quickly, you don't drift too much from a straight line on approach. But it's close enough that getting the fuel up off the surface isn't a huge pain. Also, it's convenient that I can put my lifter in a 25k orbit after takeoff and get an encounter pretty quick no matter where the station is.

Note that if you are going from Minmus to Eve or Duna, it is cheapest to go directly interplanetary, without sling-shotting past Kerbin.

If you are going to any other planet from Minmus, it is cheaper to drop into a low Kerbin slingshot.

I've been meaning to write up my tutorial on when to slingshot around the parent planet vs. when to go direct

vs. when to put the refueling station around the parent. Seems to come up a lot.

I would be interested in such a discussion. I know the fact that it is, but I don't know the reasoning.

How much Delta-V do you really save by slingshoting by Kerbin? Seems like the small amount of delta-v saved is offset by the more complected/time consuming maneuver, especially in a world were parts/fuel is free and time is the only real cost.

It's not insignificant, but you are essentially correct. I personally do it for one reason: It's fun. I also, when not going to Minmus first and launch directly from LKO to another planet, like to start early and get a highly elliptical orbit set up so right when my probe comes in to periapsis is also when it should eject to the planet, and it's moving in the correct direction so all I have to do is thrust prograde to make the transfer. It's harder, easier to screw up, and the benefits are minor but you feel like a total boss when you nail it.

[Yasmy]

Well, I've opened a can of worms before I was ready to write up the math.

From low circular orbit around the Mun, for idealized, zero plane change transfers to planets on idealized circular orbits, the delta-V cost in meters per second is approximately:

[table=width: 500]

[tr]

[td]Target[/td]

[td]Kerbin Slingshot[/td]

[td]Direct Interplanetary[/td]

[/tr]

---

[tr]

[td]Moho[/td]

[td]1135[/td]

[td]1509[/td]

[/tr]

[tr]

[td]Eve[/td]

[td]462[/td]

[td]348[/td]

[/tr]

[tr]

[td]Duna[/td]

[td]498[/td]

[td]414[/td]

[/tr]

[tr]

[td]Dres[/td]

[td]987[/td]

[td]1279[/td]

[/tr]

[tr]

[td]Jool[/td]

[td]1361[/td]

[td]1841[/td]

[/tr]

[tr]

[td]Eeloo[/td]

[td]1520[/td]

[td]2065[/td]

[/tr]

[/table]

For Mun to Eve or Duna, it's best to go direct, but for Mun to the other planets you can save a few hundred meters per second delta-V.

For low orbit around Minmus:

[table=width: 500]

[tr]

[td]Target[/td]

[td]Kerbin Slingshot[/td]

[td]Direct Interplanetary[/td]

[/tr]

---

[tr]

[td]Moho[/td]

[td]943[/td]

[td]1957[/td]

[/tr]

[tr]

[td]Eve[/td]

[td]270[/td]

[td]440[/td]

[/tr]

[tr]

[td]Duna[/td]

[td]306[/td]

[td]567[/td]

[/tr]

[tr]

[td]Dres[/td]

[td]795[/td]

[td]1700[/td]

[/tr]

[tr]

[td]Jool[/td]

[td]1169[/td]

[td]2318[/td]

[/tr]

[tr]

[td]Eeloo[/td]

[td]1328[/td]

[td]2559[/td]

[/tr]

[/table]

Thus for Minmus, you save anywhere from 170 m/s to 1200 m/s by sling-shotting around Kerbin.

One day soon I'll get around to posting the equations and derivation, and the tables for other moons. All you need to derive these values is the vis-viva equation.

Edited February 20, 2014 by Yasmy

[flyboy67109]

I'm using Kethane and mining Minmus to get fuel and O2 to sent a ship to Duna. That would make a refueling station in orbit of Minmus practical.

[Greenspan]

It's not insignificant, but you are essentially correct. I personally do it for one reason: It's fun. I also, when not going to Minmus first and launch directly from LKO to another planet, like to start early and get a highly elliptical orbit set up so right when my probe comes in to periapsis is also when it should eject to the planet, and it's moving in the correct direction so all I have to do is thrust prograde to make the transfer. It's harder, easier to screw up, and the benefits are minor but you feel like a total boss when you nail it.

Good point, I should rephase my statement to say "In a world were parts/fuel is free and time is the only real cost, and fun is the only real reward."

I use 50K as well, allthough that number was chosen completely arbitrarily.

[DMagic]

The real problem with dropping back to a low Kerbin flyby is that it is extremely difficult to set up ideal transfers this way. Everything has to be lined up correctly, and you have to take into account that it takes around 2 days (I think) to drop back down to Kerbin periapsis. There is some leeway in the positioning, of course, but if Minmus is on the wrong side of Kerbin during a transfer window then you will have to wait for another.

That said, it's still a lot of fun to do. You can also change your inclination a lot when you are out by Minmus, making transfers to Moho quite a bit cheaper.

As for the best height for a refueling station, I think somewhere around 50 - 100km is pretty good. It's easy to get to and from the surface, you can timewarp at a decent speed, and it doesn't take that much to break orbit. And don't worry about any benefit from the Oberth effect around Minmus, your orbital velocity is so low that it probably makes almost no difference.

[Superfluous J]

The real problem with dropping back to a low Kerbin flyby is that it is extremely difficult to set up ideal transfers this way. Everything has to be lined up correctly, and you have to take into account that it takes around 2 days (I think) to drop back down to Kerbin periapsis. There is some leeway in the positioning, of course, but if Minmus is on the wrong side of Kerbin during a transfer window then you will have to wait for another.

Not exactly. Minmus' orbital velocity is a meager 275m/s. You scrub, what, half of that to get your Kerbin encounter? 3/4ths? I sadly can't check now but let's say it's 3/4ths. That sounds right. That means you're scrubbing about 200m/s and end up with about 100m/s (just back-of-the-enveloping this here). If you instead scrub another 200-250m/s, you'll be going backwards in Kerbin orbit and can launch off on the opposite side. That's 2 launch windows.

Also, with this elliptic of an orbit, You're very close to Kerbin for a full half of the orbit (degree-wise). You can do your not-quite-perfectly-efficient-but-pretty-close Hohmann transfer even 45 degrees off of periapsis and still get a lot of benefit.

Essentially, what I'm saying is it's not that bad

[Kryxal]

If the Mun isn't going to cause an issue, you can also break orbit early and do an orbit and a half instead of half an orbit.

[Streetwind]

Well, I've opened a can of worms before I was ready to write up the math.

From low circular orbit around the Mun, for idealized, zero plane change transfers to planets on idealized circular orbits, the delta-V cost in meters per second is approximately:

[table=width: 500]

[tr]

[td]Target[/td]

[td]Kerbin Slingshot[/td]

[td]Direct Interplanetary[/td]

[/tr]

---

[tr]

[td]Moho[/td]

[td]1135[/td]

[td]1509[/td]

[/tr]

[tr]

[td]Eve[/td]

[td]462[/td]

[td]348[/td]

[/tr]

[tr]

[td]Duna[/td]

[td]498[/td]

[td]414[/td]

[/tr]

[tr]

[td]Dres[/td]

[td]987[/td]

[td]1279[/td]

[/tr]

[tr]

[td]Jool[/td]

[td]1361[/td]

[td]1841[/td]

[/tr]

[tr]

[td]Eeloo[/td]

[td]1520[/td]

[td]2065[/td]

[/tr]

[/table]

For Mun to Eve or Duna, it's best to go direct, but for Mun to the other planets you can save a few hundred meters per second delta-V.

For low orbit around Minmus:

[table=width: 500]

[tr]

[td]Target[/td]

[td]Kerbin Slingshot[/td]

[td]Direct Interplanetary[/td]

[/tr]

---

[tr]

[td]Moho[/td]

[td]943[/td]

[td]1957[/td]

[/tr]

[tr]

[td]Eve[/td]

[td]270[/td]

[td]440[/td]

[/tr]

[tr]

[td]Duna[/td]

[td]306[/td]

[td]567[/td]

[/tr]

[tr]

[td]Dres[/td]

[td]795[/td]

[td]1700[/td]

[/tr]

[tr]

[td]Jool[/td]

[td]1169[/td]

[td]2318[/td]

[/tr]

[tr]

[td]Eeloo[/td]

[td]1328[/td]

[td]2559[/td]

[/tr]

[/table]

Thus for Minmus, you save anywhere from 170 m/s to 1200 m/s by sling-shotting around Kerbin.

One day soon I'll get around to posting the equations and derivation, and the tables for other moons. All you need to derive these values is the vis-viva equation.

Do the values for Kerbin slingshot already incorporate the cost of leaving the Mun's/Minmus' SoI? I'm asking because I remember my last Mun lander on a 10km circular orbit required in the neighborhood of 275 m/s dV to escape and drop its Kerbin periapsis deep enough into the atmosphere to aerobrake. Thus we'd be talking maybe ca. 270 to initiate a close-pass slingshot that skims past right outside the atmosphere. That, in turn, would imply a mere ~190 m/s burn during the maneuver to make the interplanetary transfer to EVE, which seems very small to me. Not saying it can't be true, but I just want to make sure I'm reading this right.

To make this comparison even more interesting, by the way, you could add a third column with Kerbin orbit direct transfers minus an eyeballed value for getting out to the Mun/Minmus and circularizing there. This would allow a rough estimation of what you save (or pay extra) for a mission starting from a refueling depot in Mun/Minmus orbit compared to the same mission starting from a refueling depot in Kerbin orbit.

Also might be worth noting that missions starting from Minmus pay more than the theoretical figure given here because they must perform an additional plane change beyond the normal ones for most missions, as Minmus itself is in an inclined orbit.

[Yasmy]

Do the values for Kerbin slingshot already incorporate the cost of leaving the Mun's/Minmus' SoI?

Yes.

You are right that you need to consider Minmus's orbital plane, but you really don't have to worry about the delta-V. When you exit Minmus SOI on a trajectory to low Kerbin orbit, you will be going about 47 m/s relative to Kerbin. A 6 degree plane change should cost almost nothing: sqrt(2*v^2 (1 - cos(6*pi/180))) = 5 m/s.

Like DMagic said above, you can use the cheap plane change cost at Minmus to reduce the future cost of a plane correction burn in interplanetary space.

[allmappedout]

Also might be worth noting that missions starting from Minmus pay more than the theoretical figure given here because they must perform an additional plane change beyond the normal ones for most missions, as Minmus itself is in an inclined orbit.

Whilst you're right, of course, minmus is both very small and very slow, so its actual impact to your total orbital velocity (around the sun), is actually not that large (ie: you're not going to burn interplanetary and end up on a 6degree inclination, just because of minmus), plus, because it's so far out from Kerbin, a plane change maneuver correct is very cheap dV wise.

[Yasmy]

That, in turn, would imply a mere ~190 m/s burn during the maneuver to make the interplanetary transfer to EVE, which seems very small to me. Not saying it can't be true, but I just want to make sure I'm reading this right.

You got it. Kerbin's gravitational parameter is mu = 3.5316e12 m^3 / s^2.

The orbital velocity around Kerbin at 100 km is v = sqrt(mu / 700000) = 2246 m/s.

The periapsis velocity on a return from the Mun is v = sqrt(mu * (2/7e5 - 2/(1.2e7 + 7e5)) = 3088 m/s.

So there is 842 m/s of the escape from Kerbin circular orbit. Another ~190 puts you at ~1030 m/s.

Seems about right for a burn from Kerbin orbit to Eve intercept. Maybe off by 10 m/s.

[Streetwind]

You are right that you need to consider Minmus's orbital plane, but you really don't have to worry about the delta-V. When you exit Minmus SOI on a trajectory to low Kerbin orbit, you will be going about 47 m/s relative to Kerbin. A 6 degree plane change should cost almost nothing: sqrt(2*v^2 (1 - cos(6*pi/180))) = 5 m/s.

Whilst you're right, of course, minmus is both very small and very slow, so its actual impact to your total orbital velocity (around the sun), is actually not that large (ie: you're not going to burn interplanetary and end up on a 6degree inclination, just because of minmus), plus, because it's so far out from Kerbin, a plane change maneuver correct is very cheap dV wise.

I was actually thinking more from a slingshot-around-Kerbin perspective... if you happen to be dropping in from the ascending/descending nodes, you would be level with the planet, but otherwise you might come in at a significant angle.

Can you use the slingshot maneuver itself to make a plane change in addition to getting the gravity assist? Or would you have to leave Minmus' SoI in a direction more towards the ecliptic plane, and then make a midcourse correction into the final slingshot approach after you reach it?

[Yasmy]

Since you are going about 50 m/s at Minmus at Kerbin apoapsis, the maximum plane change delta-v is about 100 m/s. That's pretty cheap. So what can you get out of it in terms of a plane change relative to Kerbol? This requires just a little bit of trig.

Suppose that just before exiting Kerbin's SOI, you are on an inclined orbit, inclined by angle p₁, moving at v₁ m/s. Just outside of Kerbin's SOI, you pick up Kerbin's orbital velocity, v_K = 9284 m/s.

Then your velocity relative to Kerbol is v₂ = sqrt((v_K + v₁ cos(p₁))^2 + (v₁ sin(p₁))^2) = sqrt(v_K² + v₁² + 2 v_K v₁ cos(p₁)).

Your inclination relative to Kerbol given by is p₂ = arctan( v₁ sin(p₁) / (v_K + v₁ cos(p₁))).

Whatever your inclination was inside of Kerbin's SOI, it will be greatly suppressed outside Kerbin's SOI. If you exit Kerbin at 1000 m/s at 90 degrees inclination, your final inclination is arctan(1000/9284) = 6.1 degrees. Typically though you would want to combine your plane change burn with your transfer burn, so you wouldn't want a 90 degree exit inclination.

You may still need a second plane change burn on the line of nodes between your orbit and your target's orbit.

There are definitely delta-v savings to be found by doing part of your plane change near Kerbin. How much depends on exactly where you are going, and when. Practically, you want to invert the equations above: You know the v₂ ( = transfer v - kerbin escape v) and p₂ you need (based on the many delta-v charts, or whatever), and want to know p₁ and v₁.

(Note: if you leave Kerbin retrograde, replace v₁ with -v₁ everywhere above. This makes plane changes to the inner planets slightly cheaper than to the outer planets, per delta-V spent at Kerbin.)

[Yasmy]

Let's do an approximate calculation. It's only a loose approximation. Suppose you want to match Jool's p₂ = 1.3 degrees inclination. You need about 965 m/s in-plane exit velocity. (This is the number you read off of a delta-v chart for Kerbin to Jool transfer). This in-plane delta-v is the horizontal component of your Kerbin exit velocity: v₁ cos(p₁) = 965 m/s. Now use tan(p₂) = v₁ sin(p₁) / (v_K + v₁ cos(p₁)) to find the out of plane delta-v: v₁ sin(p₁) = tan(p₂) * (v_K + v₁ cos(p₁)) = 233 m/s.

Thus your Kerbin inclination p₁ is arctan( v₁ sin(p₁) / v₁ cos(p₁) ) = arctan(233/965) = 13.6 degrees. And your exit velocity v1 is sqrt(965^2 + 233^2) = 993 m/s. This is where you get your savings. You combine your plane change with your transfer burn and it only cost an extra 993 - 965 = 28 m/s.

To get your approximate burn delta-v at Kerbin, add about 20 m/s to this exit velocity (993 m/s). A handy delta-v map will show that it takes about 20 m/s to go from a Minmus transfer orbit to a Kerbin escape orbit. Add to this a burn at Minmus to change your inclination by 13.6 +/- 6 degrees. That will cost less than 25 m/s.

External to Kerbin, a 1.3 degree plane change would cost something below that 233 m/s, depending on your current velocity. But it would still be more than 28 m/s + the Minmus plane change delta v. (It's sqrt(2*v²*(1-cos(1.3))), which drops as you get further from Kerbin.)

Of course this was just an approximation. You may need a second inclination change burn if you are not on the line of nodes.

And someone is going to come along and tell you to just drop some maneuver nodes and play around until you got it. And I wouldn't argue with them

Edited February 21, 2014 by Yasmy

[Streetwind]

...Yeaaahhh I'm just going to smile and nod as I look at this math. Smile and nod.

[Rakaydos]

So it sounds like a high orbit at the lowest fastest timewarp is best then? (assuming your kethane lifter is optimized for payload ascent)

Or is it better to be on the bleeding edge of minmus's SoI?

Also, how much harder would it be to have a station tailing (or leading) minmus in orbit, just outside it's SoI?

[thereaverofdarkness]

I would worry less about the altitude around Miinmus and more about the direction it's going. To save dV, the orbit should be going in a direction such that you can burn into the orbit and leave Minmus' SOI on a trajectory that will feed into your Kerbin ejection burn. If you can do that, you'll save a lot more dV than you will trying to pick the best orbital altitude.

[undercoveryankee]

I would lean toward lowest fastest warp because the shorter-period orbit offers more frequent rendezvous windows.

[LethalDose]

So it sounds like a high orbit at the lowest fastest timewarp is best then? (assuming your kethane lifter is optimized for payload ascent)

Or is it better to be on the bleeding edge of minmus's SoI?

Also, how much harder would it be to have a station tailing (or leading) minmus in orbit, just outside it's SoI?

You can do this no problem. In fact, it would probably involve fewer maneuvers and would therefore be slightly more efficient. However, it's been my experience that it's easier and takes less game to rendezvous with objects in orbit of satellites vs out in empty space (i.e. very high over Kerbin). IMO, it becomes an issue of practicality (Low Minmus orbit) vs efficiency (Very high Kerbin orbit).