Kerbol polar orbit

[KAO]

challenge explains itself

is this even possible? without changing .CFG values I mean

I'll try this out soon, mods allowed.

[Bluejayek]

This is very possible, and I have done it, and even the more extreme reverse direction kerbol orbit.

One tip: Brute forcing it will take a lot more delta-V than if you use a bielliptic transfer orbit.

The principle of the bielliptic transfer is that you first burn out to a high apoapsis (100 billion meters works very well), do your orbital correction (in this case perpendicular burn to polar orbit), and then bring your apoapsis back in. The reason this works is what is called the Oberth Effect (look it up on wikipedia if you are interested). The Oberth effect essentially states that you get more energy out of your fuel when you are in a higher energy orbit (Larger orbits have higher energy). Therefore, even though you waste fuel by burning out to the high apoapsis and back in, you gain a huge amount by turning the orbit while that far out.

You can see simply how much less fuel it takes to turn to polar orbit by your velocity at apoapsis compared to at periapsis (say, Kerbin altitude). For example, the maneurver I used this for was to go into a reverse Kerbol orbit at Kerbin altitude. To do this by brute force takes about 18,000m/s of delta V (twice kerbin's velocity), which is very difficult to achieve with stock parts (The highest I have managed is around 15,000m/s). However, burning out to 100 billion meters requires about 3600m/s delta V, and the same to return. At apoapsis your velocity is ~1000m/s, and so it takes 2000m/s of delta V to reverse the orbit. Adding these up, the total delta V for the bielliptic maneurver is about 9200m/s, just about half of what it takes by brute force.

For the polar orbit case, the most efficient brute force burn (single burn at 135 degrees to orbit plane) requires ~13,000m/s of delta V. Doing the same adjustment at 100 billion meters requires ~1500m/s of delta V. After acounting for the two transfer burns, the bielliptic method requires 8700m/s of delta V while the brute force method requires 13,000m/s, a gain of ~4300m/s.

Bottom line, use the bielliptic transfer for most non time critical missions in kerbol SOI!

Edited August 31, 2012 by Bluejayek

[Bluejayek]

Below is a graph showing the delta V required to attain a polar orbit at kerbins altitude as a function of the apoapsis of the bielliptic transfer. As you can see, 100Gm is around where you hit diminishing returns. The second graph shows why you dont go out as far as possible. That is the time in real life minutes you have to wait while at 100,000x time warp as a function of apoapsis altitude.

Below pictures show me attaining the polar orbit using the described bielliptic transfer out to 100Gm. I was easily within fuel budget, and in fact jetisonned about 8000m/s Delta V worth of fuel when the kraken was being annoying.

Oh, just a note about these delta V requirements. These are over and above whatever it takes to escape kerbins SOI, which I do not have a good handle on. If somebody could enlighten me that would be great.

Edited September 1, 2012 by Bluejayek

[MaverickSawyer]

Could you use a gravity slingshot at the Mun or Minmus to reduce the dV required? I had no trouble shifting from a equatorial approach to a polar approach with a Mid-Course Correction (MCC). maybe fire a big-ass booster at periapsis at the Mun? I'll have to give this a try.

[Shpaget]

I got to a 200 km circular polar orbit with a direct burn in the general direction of North, achieved a temporary orbit of 90-190, pushed the inclination to 90° and raised both Ap and Pe to 200, in 42 minutes or less, mission time.

http://i.imgur.com/z5B70.png

I did it using something like this:

http://i.imgur.com/trLTz.jpg

I still have plenty of fuel.

http://i.imgur.com/9SRJG.png

[Bluejayek]

Nice ship shpaget, but the challenge was KerbOL oorbit (sun) not kerbin

[Shpaget]

Oh. I knew I was missing some crucial part. It just seemed too easy.

[sleedog1]

when i first started to play ksp i couldn't get into a normal orbit i always ended up in a polor orbit it is all in the way you take off

[vXSovereignXv]

I have an idea on how to do this: http://en.wikipedia.org/wiki/Ulysses_%28spacecraft%29. My idea is to take a probe on a 53 day orbit and fly by kerbin after two orbit. I should in theory be able to use a gravity assist to do the plane change. It's getting late now so I'll try it tomorrow.

[MaverickSawyer]

Well thought out plan. I like the odds of it working. Just be gentle with the thrust...

[vXSovereignXv]

I tried the gravity assist. It didn't work as well as I hoped. Best I got was an incline change of about 20 degrees. Maybe with several passes or a larger planet I could get there.

[MaverickSawyer]

I tried the gravity assist. It didn't work as well as I hoped. Best I got was an incline change of about 20 degrees. Maybe with several passes or a larger planet I could get there.

-snip-

Hmm. Use a spaceplane and an aerodynamics-boosted plane change? I know it's theoretically possible, especially with the lack of reentry heating and such... but can you run a scramjet for a quarter of the way around Kerbin at a power level to maintain Kerbol-orbiting velocity AND pitch "down" or "up" enough to pull it off?

BTW, what was the encounter velocity?

[vXSovereignXv]

I was going 10800 m/s relative to Kerbol just before the Kerbin encounter. At kerbin perigee I was going 3900 m/s relative to Kerbin. I think an improved angle of attack would help, but it'd still take multiple assists to cancel out the equatorial velocity. I'm not sure about the space plane idea. They don't seem to offer much lift until well into the atmosphere where the drag would kill most of the speed.

[MaverickSawyer]

True. Maybe around 45 km? That's above when Vlad's airbrakes begin to have any real effect, which I have found is an indicator of serious drag approaching. Also, a Scramjet might help maintain velocity, although, at almost twice orbital velocity, it'll have to be crewed by Jeb.