Question about Normal/Antinormal burns in elliptical orbits

[EndOfTheEarth]

Okay, physics people, this one's for you.

Background: I've got a fuel tug that just entered orbit around Laythe. The orbit, for the moment, is fairly epiptical, and at about a 45 degree inclination to the equator. The objective is to dock with a lander that has just come back from a surface sortie, and is now sitting in a low, circular, equatorial orbit.

Picture for clarity:

Naturally, the idea here is to save fuel instead of spending it, so my question is: When modifying the orbital inclination of my tug to match the lander, would I be using less fuel at the ascending node, descending node, or would the resultant burn time for either node look about the same?

Bonus: If I want to lower my apogee at the same time, does this change the answer?

Additional stuff, if it helps:

Stock game, no mods, most recent version, I think.

Thrust on the tug is provided by four LV-Ns, and a handful of RCS thrusters for close-in docking.

The ship is currently holding 3014LF/3683Ox and 596 Monoprop, arranged in four T800s, one S3-7200, and four radial-mounted Cylinder monopropellant tanks.

My rough math says that this plus extras like the docking ports, ASAS, RTGs, and so on all translates to a wet weight of 49.87 tons, if the wiki is accurate.

EDIT More numbers:

Lander (Circular-ish orbit):

-Apogee: 62,917

-Perigee: 58,649

Fuel Tug (eliptical orbit)

-Apogee: 789,610

-Perigee: 62,728

Other Data:

Inclination difference: 41.7 degrees

DN Altitude: 279km

AN Altitude: 288km

AN/DN Altitudes are approximations based on the staging view altitude indicator while passing though each, as the ship is gaining/losing altitude at a rate of about 1km/s by that point in the orbit.

Edited December 15, 2014 by EndOfTheEarth

[GoSlash27]

Very interesting problem!

I would think your burn would be the same either way because your velocity would be the same.

Interesting concept to look at: It may be cheaper to raise your periapsis first, then correct your inclination and finish your transfer. You might want to check that with some maneuver nodes.

You could also see what it costs to do a radial in burn at the ascending node until your apoapsis is at the descending node, and then correct your inclination at the new apoapsis.

I could be totally wrong, so I'm interested to hear what the others have to say about it.

Best,

-Slashy

[Red Iron Crown]

From the look of your orbit, AN occurs at higher altitude and thus lower speed, so a direct inclination change will be cheaper there (you could confirm by setting a node at each and seeing the dV cost). Hard to tell from the angle of you pics but it looks like about a 45 degree inclination change, not enough that I'd bother with a three burn strategy (raise Ap way up from DN, inclination change at AN, drop Ap back at DN).

Lowering your Ap at the same time won't be most efficient. Better to do your inclination matching, then at Ap lower your Pe to match the target orbit, then at Pe burn to get your rendezvous.

[Reactordrone]

Plane changes are cheaper at lower speeds so it might pay to raise your apoapsis by burning prograde at either the AN or DN (whichever is closer to the planet) then do the plane change and aerobrake back down to the correct orbital altitude.

[LethalDose]

Very interesting problem!

I would think your burn would be the same either way because your velocity would be the same. [...]

I could be totally wrong, so I'm interested to hear what the others have to say about it.

Yep, you're wrong.

(PS That's not an attack, it's simply a statement of fact)

In this case only making a plane change would be cheaper at the DN node because the DN node is closer to the AP (I could have paralax error, but I don't think so).

However, if RIC is correct (The AN is higher than the DN because the AN is closer to the AP), then, again the AN would be the cheaper place. I think it's pretty close

So in general, you want to burn to correct inclinations at the node closest to the AP for reasons RIC stated: Nodes near the AP have higher altitudes and lower velocities, making inclination changes cheaper.

Sometimes its cheaper to move the the AP towards a node with extra burns, sometimes it isn't. You're kind of in a worst case scenario here with the axis of the inclination being nearly perpendicular to the semi-major axis (the AP-PE axis).

If you want to try to save fuel on this rendezvous, I would make the following three burns in order:

1. Burn prograde and radial out at the AN Until the AP lays on top of the DN.
   
2. Burn retrograde and normal at AP to fix inclination (I get normal and anti-normal mixed up, just do the right one) and set the PE to your target's altitude (target's orbit looks pretty circular)
   
3. Burn retrograde at PE until you get an intercept in an orbit or two (this is basically just burning into breaking orbit to get synchronized with the target)
   

Then just rendezvous when you get that encounter. This is basically a bi-elliptic transfer, adjusted for the specific situation. I think it'll be the most efficient way to rendezvous in this case, but I'm really not sure how much it's gonna save you. Bi-elliptics can be more efficient than Hohmanns in the right circumstances, which occur when you have very eccentric orbits. Hence, I want to say "the higher the better" on step one, but make sure you don't escape from Laythe.

Addendum: This is all derived from the equation for dV cost for maneuvers with inclination changes:

dV = (V₁² + V₁² - 2V₁V₂cos θ)^1/2 (eq 4.74)

One of the implications of the equation is that lower speeds lead to lower dV costs for inclination changes.

Edited December 13, 2014 by LethalDose

[purpletarget]

With only 45 degrees to change, as was already mentioned, there isn't any savings to be had by purposely raising the Ap.

Also you have a circle within an ellipse, and the Major axis appears to be at approximate right angles to the An/Dn nodes. As a result, there isn't likely to be an appreciable difference between which node you use to make your inclination change.

If the Major Axis was aligned closely, (Ap/Pe close to An/Dn) then you would want to do the burn at whichever was closer to the Ap for some marginal savings.

Secondly, yes, control your Ap at the same time. If you do a normal/antinormal only to adjust your inclination, the nodes by default will also increase your orbital velocity vector at the end, so chances are your inclination burn location will become a new Pe, and you'll have to make up that dV a second time lowering a likely higher Ap. When you plan out your burn, use a combination of Normal/Antinormal (whichever is helping the inclination) with some retrograde. You can probably manage to get both the inclination you need plus getting your Pe down to the target altitude in a single burn, and that should save you at least some juice.

ETA: Given your numbers you have enough dV in that thing to get off Kerbin almost twice (TWR notwithstanding), ~8km/s+...so it's not like saving 10m/s here and there is going to affect you too much. Also, a really quick way to be certain of you mass at any time in flight, would be to check the Info button for your spacecraft in the map view. From that, and your fuel numbers, you could get an almost exact dV very easily.

Edited December 13, 2014 by purpletarget

[LethalDose]

@OP If you can give us the 5 parameters listed below, we could get some best and worst case scenarios.

• Target PE & AP
  
• Vessel PE & AP
  
• Inclination change
  

If you can give us the argument of the ascending node and and argument of the PE, we could do it exactly. If you can't get these numbers, then the approximate altitude at the AN or DN will do in a pinch.

I'd actually like to crunch the numbers.

Edited December 13, 2014 by LethalDose

[GoSlash27]

LD,

Not taken as such. Neat resource and thanks.

Best,

-Slashy

[Empiro]

As others have said, inclination changes are cheaper if you're moving more slowly (usually means at a higher altitude). However, if you're able to combine maneuvers, then that usually leads to even greater savings (think of a triangle).

For example, if your AN/DN was instead right at the PE, which touches the orbit you're trying to match, then doing a normal burn combined with a retrograde burn at the same time at PE might lead to greater savings even though it's more expensive to do the inclination change there.

In your particular example, I think the cheapest would be to do the following:

-At either the AN or DN (try both to see which is cheapest), add a maneuver node with enough of a normal component to match inclination. Because your inclination change is pretty large, you'll want to add a radial / retrograde component to make it so that your PE is within the atmosphere. Play around with adding more / less of each and see what would be the cheapest. You don't care what your AP is, just that your PE is in the atmosphere.

-Then allow yourself to aerobrake until your AP matches the target orbit

-Raise your PE to match the target orbit.

Edited December 13, 2014 by Empiro

[Kryxal]

Another vote for aerobraking here, I was about to suggest it when I saw the last post...

[EndOfTheEarth]

@OP If you can give us the 5 parameters listed below, we could get some best and worst case scenarios.

• Target PE & AP
  
• Vessel PE & AP
  
• Inclination change
  

If you can give us the argument of the ascending node and and argument of the PE, we could do it exactly. If you can't get these numbers, then the approximate altitude at the AN or DN will do in a pinch.

I'd actually like to crunch the numbers.

Okay then! More numbers:

Lander (Circular-ish orbit):

-Apogee: 62,917

-Perigee: 58,649

Fuel Tug (eliptical orbit)

-Apogee: 789,610

-Perigee: 62,728

Other Data:

Inclination difference: 41.7 degrees

DN Altitude: 279km

AN Altitude: 288km

AN/DN Altitudes are approximations based on the staging view altitude indicator while passing though each, as the ship is gaining/losing altitude at a rate of about 1km/s by that point in the orbit.

Thanks to everyone for their help so far! I haven't touched it yet, because I'm curious about what you folks come up with.

[LethalDose]

Okay then! More numbers:

Lander (Circular-ish orbit):

-Apogee: 62,917

-Perigee: 58,649

Fuel Tug (eliptical orbit)

-Apogee: 789,610

-Perigee: 62,728

Other Data:

Inclination difference: 41.7 degrees

DN Altitude: 279km

AN Altitude: 288km

AN/DN Altitudes are approximations based on the staging view altitude indicator while passing though each, as the ship is gaining/losing altitude at a rate of about 1km/s by that point in the orbit.

Thanks to everyone for their help so far! I haven't touched it yet, because I'm curious about what you folks come up with.

Wow... if there's only ~ 9 km difference in alt between your AN and DN, it doesn't really matter which one you would burn at.

Thanks for the data (mmmmm data), I'll get to these when I have a chance.

[LethalDose]

Okay, ran the calculations. The results are below:

Bi-elliptic transfer, 3 burns, as described in my post

• Burn 1: Made at DN to Boost AP to 3,000 km altitude at AN, 759.9 m/s Prograde and radial
  
• Burn 2: Made at transfer AP for inclination change and target orbit, 599.1 305.7 m/s retrograde and normal
  
• Burn 3: Made at transfer PE for circularization at ~60 km for rendezvous, 585.4 m/s retrograde only
  

Total dV for bielliptic transfer: 1944.4 1651.1 m/s

Direct transfer, 2 burns, as typical for a Hohmann transfer

• Burn 1: Made at AN for inclination change and target orbit, 2233.6 1245.3 m/s retrograde and normal
  
• Burn 2: Made at transfer PE for circularization at ~60 km for rendezvous, 151.5 m/s retrograde only
  

Total dV for direct transfer: 2385.1 1396.8 m/s

Bi-elliptic Savings: 440.7 (18.5% savings over direct)

Difference: 254.3 m/s in Favor of direct transfer (19.7% increase in cost)

The values above assume "rendezvous" is a circular orbit with a semi-major axis of 560.7 km (60.7 km above sea level, equivalent to the lander's reported semi-major axis). Personally, I would perform the final burn in two parts, the first synchronize a good good target intercept, the second to finish circularization at the target.

Going higher improves dV savings; 3,000 km altitude was chosen to be high, but safely with Laythe's SoI (approx 3.2 Mm). Only going to an altitude of 2000 km leads to a total dV cost of 2054.7 m/s (330.4 m/s savings, 13.9% savings) 1651 m/s. The break-even point for the maneuvers occurs when the bi-elliptical transfer's apoapsis is approximately 850 km above Laythe's sea level. There does not appear a break even point, though break even points appear at higher inclination changes.

So... Go bi-elliptic transfer. freakin' a... Values have been corrected for calculation error on my part. Original values have been struck for transparency.

Addendum: I'm not sure how people are discussing using an aerobrake for this, but I think using the aerobrake for third burn of the bi-elliptic transfer could help.

Edited December 15, 2014 by LethalDose

[EndOfTheEarth]

Okay, ran the calculations. The results are below:

Bi-elliptic transfer, 3 burns, as described in my post

• Burn 1: Made at DN to Boost AP to 3,000 km altitude at AN, 759.9 m/s Prograde and radial
  
• Burn 2: Made at transfer AP for inclination change and target orbit, 599.1 m/s retrograde and normal
  
• Burn 3: Made at transfer PE for circularization at ~60 km for rendezvous, 585.4 m/s retrograde only
  

Total dV for bielliptic transfer: 1944.4 m/s

So... Go bi-elliptic transfer.

Addendum: I'm not sure how people are discussing using an aerobrake for this, but I think using the aerobrake for third burn of the bi-elliptic transfer could help.

Nice! I'll give this a shot next time I'm on. Thanks for all the number crunching!

[Kryxal]

It might be worth seeing how well raising your PE first would work instead ... I have some craft in highly elliptical orbits where spending 100 m/s at AP leads to large savings on the plane-change.

[LethalDose]

It might be worth seeing how well raising your PE first would work instead ... I have some craft in highly elliptical orbits where spending 100 m/s at AP leads to large savings on the plane-change.

I'm not sure which PE you're talking about, but I can't think of any situation where raising PE is going something you want to do. In the original situation, the original PE is already above the target PE, so you want to lower that. Any situation where you want to be changing inclination you want to be done as high and as slow as possible, which means the lowest possible PE. Remember, if you want to raise your PE, you accelerate at AP. You don't want to accelerate at the AP if you want to go slow at the AP...

I don't know what direction you were spending this magical "100 m/s at AP", but if it was in the prograde direction, I guarantee wasn't saving you dV for inclination change.

I take that back. Burning prograde at AP will raise the orbit at all points (except for the AP, of course), however, that 100 m/s will always be better spent at the AN or DN so it raises the opposing node as high as possible. Specifically, it's best spent at the lower node to raise the higher node even higher.

Edited December 15, 2014 by LethalDose

[LethalDose]

Two additional addenda:

First, for clarity, when I said "aerobraking could help", I'm indicating there is likely a dV savings to be had even thought he second burn would be greater (to push the Pe into the atmosphere). And really, the overall dV cost would be more with aerobraking, it's just that you're getting some of the dV for "free" from the atmosphere. I don't think the savings using an aerobrake at that point would really exceed 400 m/s, and thats being pretty generous. But every bit helps, I guess.

Second, the numbers for the direct transfer are done assuming that the point of the first burn would become the AP for the transfer... The more I think about it, the more I think that's a crap assumption because if you burn retrograde at the node, there's always going to be some radial component to the velocity (unless orbital v=0). Therefore the numbers provided for the direct transfer are actually lower than they should be since the actual transfer orbit would have a longer semi-major axis leading to a higher periapsis velocity and greater dV cost to rendezvous. I don't think the difference is of a magnitude that is relevant, but it doesn't me that the savings presented is actually more of a lower bound; the actual dV savings of the bi-elliptic is likely a bit higher than present.

The dV values have been corrected for the non-zero radial velocity. Adds ~ 100 dV

Edited December 15, 2014 by LethalDose

[GoSlash27]

Hmm...

Bi-elliptic transfer is actually preferable after all...

I hadn't run the numbers (although this post has inspired me to start doing that). It was just a hunch.

Best,

-Slashy

[LethalDose]

Well... Crap. This is embarrassing.

I found an error in my equations. Specifically, I forgot to multiply the 2Cos(theta) term by the product of the pre- and post-burn velocities in dV calculations

[long stream of foul language]

The post above has been corrected. The Bi-elliptical is NOT more efficient.

I apologize for any inconvenience.

Edited December 15, 2014 by LethalDose

[Empiro]

Also just as a quick addendum:

I went into the game and tested this out via hyperedit, and it looks like LethalDose's new numbers are pretty accurate. You can squeeze and save a bit more by burning a bit more retrograde on your initial maneuver so that your PE is within the atmosphere and aerobrake until your AP matches your target.

[LethalDose]

Also just as a quick addendum:

I went into the game and tested this out via hyperedit, and it looks like LethalDose's new numbers are pretty accurate. You can squeeze and save a bit more by burning a bit more retrograde on your initial maneuver so that your PE is within the atmosphere and aerobrake until your AP matches your target.

Thanks for the check on the numbers and I'm glad to hear that they pan out.

It's worth noting that the numbers presented don't represent the most efficient possible maneuvers. Doing that would likely require some pretty freaking advanced math (differential equations and iterative numerical solving methods). I'm not going through that crap because I don't expect the numbers would change that much (seriously, single digit differences) and no one would be able to actually perform the burns that precisely anyway.

[EndOfTheEarth]

Hi there! Just checking in to say that I made the burns using the direct transfer method with aerobreaking, and the d/V required was indeed around 1350m/s. I still have plenty of fuel to spare for docking, so it looks like my crew might finally get home from Laythe! Thanks again!