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180 Degree Inclination Orbit


Chads
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Yes, launch to the West. 180° is an equatorial orbit, only opposite the usual way. Remember to pack a little bit more dV since you won't be using Kerbin's rotational velocity, but rather be fighting against it.

Edited by Gaarst
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19 minutes ago, klesh said:

You don't need mech jeb to fly due West.  You need only the A button on your keyboard.

In his defence: he didn't said it is necessary. 

Didn't t even claimed to be related to the question. 

 

@CrisCopper Yes,  MJ is useful.  But many people prefer to not use autopilot.  Also,  basic understanding of orbital maneuvers and parameters are useful with or without MJ

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If the orbit's out at about 8000km it'll cost about the same to launch East and reverse orbit at apoapsis as it would to launch west. If it's a higher orbit than that you might actually save some delta-v requirement launching East rather than doing a direct Westerly launch.

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6 hours ago, Reactordrone said:

If the orbit's out at about 8000km it'll cost about the same to launch East and reverse orbit at apoapsis as it would to launch west. If it's a higher orbit than that you might actually save some delta-v requirement launching East rather than doing a direct Westerly launch.

I see.. Because the "head start"  you got launching east and the low cost to invert the high  orbit. While I know the concept behind it never took the time to know  the exact "turning point"  (forgive the attempted pun).  Did you considered the  orbit  reversal at target altitude or raise it further higher (edge of SoI) for even  cheaper orbit reversal? 

 

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11 hours ago, Reactordrone said:

If the orbit's out at about 8000km it'll cost about the same to launch East and reverse orbit at apoapsis as it would to launch west. If it's a higher orbit than that you might actually save some delta-v requirement launching East rather than doing a direct Westerly launch.

It kinda depends on how inefficient the launch is though - back in the day with 'thicker' atmo, a much steeper (theoretically inefficient but the best possible with drag + heating) launch profile was the way to go - this would give the perfect profile for reversing at apoapsis ~8000km because the horizontal velocity remaining for a steep ascent was low enough, i.e. ~175m/s before circularisation burn. The current atmosphere is pretty realistic so more horizontal profiles (20-35 degrees) starting from 25-35km (like SpaceX's) can save an extra 250-350dV which also leaves a much higher horizontal velocity for the same altitude so we'd need to go higher before it becomes optimal.

So strangely enough, if Chad is normally inefficient with his launches, Reactordrone's method is the way to go. Otherwise, launch West. 

There must be some Math-fu to work out what's best because launching straight up to 8000km and letting gravity stop you there, then going west - would that be even more efficient?

 

 

 

Edited by Weywot8
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1 hour ago, Weywot8 said:

Snip

I think you missed @Reactordrone point.

Launching to east is cheaper than to west since we got that "head start" from Kerbin's rotation. Assuming a infinite thrust and no drag whatsoever that difference would be the cost for reversing the actual trajectory or twice the Rotation velocity (~350m/s). In reality because of higher drag and gravity losses, the difference is even higher, but for the sake of the discussion lets just assume a perfect trajectory.

Two rockets, identical at the launch time, are launched in opposite direction, both end up in circular orbit at 75km. At this point East Rocket have 350m/s higher deltaV budget. Now West Rocket need to raise to expend m/s to raise it's orbit apoapsis while East Rocket will need expend (x+y) m/s, where y is the cost to reverse trajectory, to reach the same orbit. For some target orbit y=350m/s, lower than that West Rocket need less deltaV, higher and is East Rocket that have the upper hand.

Assuming that unrealistic optimized situation I find the critical altitude to be 11 500Km, conveniently about the orbit of the Mun. I'm courious about Reactordrone 8 000km ballpark, since it probably assume a more realistic set of parameters.

 

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Hmm

5 hours ago, Spricigo said:

snip, snip

 

urm, well, I supposed I assumed  @Reactordrone knew what he was talking about and work things out from there to the only logical scenario where what he said was true . I'll admit probably jumping the gun with "if @bewing likes it, it has to be correct" (at least in some form or another, right?:sticktongue:)   

----> going by your logic instead, for reversal at 350m/s budget or less simply means altitude where orbital velocity = 350/2 = 175m/s and anything above that. A bit of a rehash of the plane change at infinity = 0dV problem/solution.

In that case, you might have converted km vs. meters wrong somewhere - because the Mun's orbital velocity is ~550m/s, Minimus'  is ~275m/s...so  almost halving it again to 175m/s should put it  'out there somewhere' with an SMA ~4x that of Minimus: certainly not in Kerbin's SOI. Unless I've seriously mixed up the numbers or missed the point again. 

Let me know if  get something else by a different method. :cool: But it's all cool otherwise.

Edited by Weywot8
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Yeah, slight miscalculation. Break even is around about the Muns orbit or just under 12,000km so anything that has to be between the Mun and Minmus you'd be better off launching east.

Having said that it is quite cheap to raise your apoapsis from that point (and essentially free to lower it again afterwards using aerobraking) so you may still be better off with an eastward launch for orbits of around 8000km. For example, raising your apoapsis from 8000km to 15000km costs about 50m/s and reversal there will cost about 265m/s for a total of 315m/s which is lower than the 350m/s extra that you spend to launch Westwards.

If you don't need an orbit that high it's obviously not worth spending 830m/s pushing up to 8000km in the first place.

Edited by Reactordrone
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Ah, not circular orbit then, I see where I went wrong now . But then, why enter a 72x8000(or 11,500)km orbit in the first place if we are going to reverse in the end?

I had assumed Reactordrone's logic was maximum efficiency by reversing at the Ap of your launch trajectory before circularisation - or for the more orbit minded, in an elliptical 'orbit' with a -ve Pe given that the 'parabolic' surface trajectory of a launch is actually an ellipse with the center of Kerbin as one focus. Same reasoning applies, just not constraining Pe to +ve values. 

Edited by Weywot8
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 @Reactordrone I thought the 8Mm was a value you got from practical experiment or at least a estimative of typical deltaV losses (launching east VS west). Unfortunate if that is not the case. 

2 hours ago, Weywot8 said:

Ah, not circular orbit then, I see where I went wrong now . But then, why enter a 72x8000(or 11,500)km orbit in the first place if we are going to reverse in the end?

Because we need to reverse the trajectory somewhere(we start with some velocity toward east) .  Given that to reverse the trajectory we need to expend twice the velocity we have when performing the reversal burn,  the ideal moment is at the apoapsis (lower velocity) .  Doing it at the launch time is the only possible exception because we didn't reached full orbital velocity yet. 

There is also the possibility of a bi-elliptic transfer save bit extra m/s (for the discussion we have there just a detail but one deserve some tinkering which I didn't give)  but the General idea is: if the target orbit is high enough,  launching east is cheaper. 

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:D Guessing we are still on different wavelengths - to clarify for @Spricigo, Why enter a 72x11,500km orbit, effectively 'circularising' at 72km first then boosting to 11,500km Ap when we can skip the 'circularisation' and just launch a rocket from the surface to give us an elliptical 'orbit' of dimension of  [some negative number] x [some number smaller than 11,500, say 8000] km, that puts the orbital velocity at 175m/s at Apo?  The 'circularisation' step seems a bit arbitrary as we will be reversing the orbit anyway (Pe of 72 km goes down to -600 km(?) before going back up again) - this whole circularisation is what threw me off and then got misapplied at various points), and being able to rescue/plan launches destined for lower orbits makes this sort of maneuver increasingly useful.

Said another way, if we were planning to do this reversal all along, why waste dV 'circularising' at 72km the wrong way when we would spend an equal amount reversing that velocity as well, forcing the effective altitude for efficient reversal all the way out to 11,500km?

  • this 'reverse direction' maneuver using ~350m/s is doable and I've actually done it at altitudes much lower than 8000 km (there was a retrograde KEO/GEO contract) as part of the whole "Oops oh *&%!...OK, well that worked out", hence assumed this was what @Reactordrone was referring to. Thought perhaps Reactordrone had worked out 8000km as a sweet spot for doing this.

So in figuring out how useful this can be, without raising Ap above final orbit, the follow on question is how low can Ap get for this to still work for Kerbin, which I assumed required an additional constraint of how inefficient/efficient (steep) one normally launches - otherwise it probably degenerates on one extreme into the bi-elliptic transfer for maximum savings on the plane change (agree with you there) and launching directly West at the other extreme.  Also Kerbin's atmosphere tends to force retrograde launches into a steeper, less efficient ascent profiles to safely 'add' 350m/s dV to the launch without blowing up/excessive drag, so the penalty is more than 350m/s dV, unless the prograde launch was pretty steep to begin with.

The actual problem then becomes an iterative numerical exercise, optimizing for the lowest orbital height where reversal dV= (the most efficient retrograde launch profile dV)-(most efficient prograde launch profile dV) to that lowest orbital height but Jeb would go :huh: and just fly the damn thing with MOAR boosters.

Edited by Weywot8
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I see... 

72km is just a arbitrary value (and I fact my calculation was with 75km and rounding induced errors) .  Assuming we first reach a arbitrary circular orbit just out of the atmosphere is not necessary but useful.

There here is several factors that only matter during launch/atmospheric.  All those factors contribute to make launch to East cheaper (possible exception being inconsistent Piloting).  Assuming we first reach a arbitrary circular orbit just out of the atmosphere let us recognise there is a difference while for a first approximation is sufficient to know that launch we st will use some extra deltaV ,  at minimum twice Kerbin's rotation velocity. 

Even if we directly launch (which indeed will increase overall efficiency)  we need,  in actual flight, time perform the maneuver.  In fact we don't 'skip' any flight phase,  rather we perform then in a continuous and combined way.  We just pass from putting effort in 'getting into space'  to 'staying out of the atmosphere'  to 'reaching target apoapsis'  without interruptions or much a notice. In any case the later phases will be performed at higher altitude then the previous (just because we are going up).  For the sake of analysis is useful to consider its done at a certain altitude,  by the very nature of our procedures a arbitrary altitude. 

We may not know exactly where the turning point is, but  Reactordrone's conclusion that if the target orbit launching east is a better option.  

And for a target orbit at least that high the most fuel-efficient way it's to raise the apoapsis to the edge of the SoI,  invert the orbit, adjust apoapsis/periapsis to target orbit. 

 

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