UnHohmann transfers

[GoSlash27]

4 hours ago, steuben said:

citation or proof required.

steuben,

 Actually, no it's not. All that is "required" is to acknowledge that you're no longer dealing with a Hohmann transfer at that point, so there's no upper bound.

Best,
-Slashy

[Pand5461]

Well, there will be an upper bound realistically because rockets cannot accelerate instantly (google "brachistochrone" and "torch drive"). But it is ridiculously high to the point it's absolutely impractical gameplay-wise.

There might also be a minimax value, meaning that we can calculate minimum deltaV for transfer initiated on specific date and choose the date that has maximum value of such minimum. That deltaV would guarantee that it is possible to reach your destination starting whenever you like but still leaves a room for making mistake you won't be able to recover from.

[Spricigo]

Ok, lets try another take. At least we agree that the for minimal deltaV expenditure we need to do a Hohmann transfer:

So what we want is (1) go from the lower to the higher orbit and (2) match the position with a planet in the higher orbit. For simplicity lets assume both orbits are circular and ignore the (Oberth) effect from the gravity of the planets.

The cost of (1) is the cost of a Hohmann transfer, a given amount at Pe to raise the Ap, and some more at Ap to raise the Pe.  Any extra and we will overshot.

So the question is: what is the cost of (2) ?In terms of deltaV it cost nothing. We just need to stay in the lower orbit until we catch up with the planet. So,  that is what we do unless we have some time constraint.

 

Notice, we can just wait for the launch window or we can launch now and wait a number of orbits until we meet the planet at the apoapsis. In both cases the required deltaV will be the same.

 

 

 

[GoSlash27]

25 minutes ago, Spricigo said:

 

^ This, and I want to stress that anything that does not look like this is not a Hohmann transfer. If this is what the OP is considering, then we can establish a calculable upper bound for the DV required. If it's not, then there is no upper bound.

Best,
-Slashy

[Capt. Hunt]

On 1/19/2018 at 10:50 AM, steuben said:

I'm looking for the max delta-v required to go from orbit to orbit, assuming a Hohmann style transfer. So one burn at each end, and maybe one in the middle for plane change.

The math is looking pretty dense for me to take a run at it myself. And a lot of what I'm seeming to find just calculates the Hohmann Δv. Or I'm not reading it very well... also probably true.

Anybody got some pointers? Absolute preference is a nice precalculated table. But, I'm kind of prepared to do the lifting needed to build it... I just can't seem to find a decent place to start on the calcs.

Saying you want to take the maximum possible transfer delta-v without a brachiostrone is nonsense.  Hohmann style transfers specifically are orbits that take full advantage of oberth effect to get optimal low Delta-v burns, while a brachiostrone is a transfer orbit with the maximum possible delta-v you can use.   Essentially, Brachiostrone delta-v really can be infinite, the only limiting factors are the Isp of your rocket engine and your fuel supply. 

Transfer orbits are a spectrum, from an ultra efficient Hohmann to a max burn Brachiostrone.  The question becomes where on this spectrum do you want to be?

 

[Snark]

On 1/24/2018 at 3:00 PM, Capt. Hunt said:

Hohmann style transfers specifically are orbits that take full advantage of oberth effect to get optimal low Delta-v burns

Actually, Hohmann transfers aren't particularly related to Oberth at all-- what they're taking advantage of is geometry, specifically that all maneuvers are either perfectly or relative to orbital velocity around the primary, and thereby maximize dV efficiency.

 

[OhioBob]

13 hours ago, Snark said:

Actually, Hohmann transfers aren't particularly related to Oberth at all-- what they're taking advantage of is geometry, specifically that all maneuvers are either perfectly or relative to orbital velocity around the primary, and thereby maximize dV efficiency.

^^^ This.  The distinguishing characteristic of a Hohmann transfer is that the burns are performed at periapsis and apoapsis of the transfer orbit were the velocity vectors of the transfer orbit are tangent to the initial and final orbits.  For both burns, the initial and final velocities have different magnitudes but the same direction.  The fact that no change of direction is necessary is what makes a Hohmann transfer so efficient.  Oberth effect has nothing to do with it.

For a non-Hohmann transfer, at least one of the crossing points between the transfer orbit and the initial and final orbits is at an angle (usually where the transfer orbit crosses the final orbit).  So now the burn requires not only a change in magnitude but also a change in direction.  Also note that a non-Hohmann transfer requires that the transfer ellipse be larger, so that makes the delta-v greater to begin with.  But the big reason why the delta-v is higher for a non-Hohmann transfer is the angle at which the orbits cross.

You can of think of it this way, if one car traveling 100 kph is approaching another car traveling at 80 kph from behind, then the approaching car must slow down only 20 kph to avoid a collision and match the other car's speed.  However, if those two cars are coming together at a right angle intersection with one car crossing the path of the other, then the relative speed between them is 128 kph.

The following helps to illustrate the difference between Hohmann (Figure 4.11) and non-Hohmann (Figure 4.12) transfers.  Figure 4.12 is an example of what is called a "one-tangent burn".

 

Edited January 26, 2018 by OhioBob

[Capt. Hunt]

ok, so I guess I was a little confused on that point, but my main argument still stands, what the OP wants is nonsensical.

[Spricigo]

1 minute ago, Capt. Hunt said:

 what the OP wants is nonsensical.

or rather he just didn't find a way to explain what he want.

His analyses seem to be flawed. But that may be a case of "lost in translation".

 

My question:

Is his analysis based on incorrect assumptions or is he using a incorrect terminology to explain it for us? For me it seems to be a mix of both, but without @steuben's help we cannot answer that.

[Tig]

I think this is a little lost in translation and I think the word that is throwing everyone off, including me, is "maximum"

I hate to put words in the OP's mouth, but I think he is asking the following:

In the set of all future optimum hohmann transfers between Planet A and Planet B what is the worst case scenario for the delta v required?  In other words his question inherently involves non-circular orbits/orbits with inclination changes.  (For circular orbits without inclination changes the set of delta v required for all future hohmann transfers between A and B would all be equal, of course)  For instance, each year there is some optimal minimum delta v required to go from low orbit Kerbin to low orbit Eeloo.  Create an array with the index representing the year and the value equaling the optimal minimum delta v that year to transfer from low orbit Kerbin to low orbit Eeloo.  The theoretical array has an index from 1 to infinity. What is the "maximum" number in said array?

 

I think that's what the OP is asking.  If its not, well, I still think my version is somewhat interesting...   (at least its solvable, lol)

 

PS.  I know the optimal transfer window from Kerbin to Eeloo is not exactly one year apart.  But dammit, its close enough to make my point.

[Kryxal]

Each is a local minimum of delta-v, but these minima are not equally good, so it's basically the maximum local minimum (I think).

[GoSlash27]

35 minutes ago, Kryxal said:

Each is a local minimum of delta-v, but these minima are not equally good, so it's basically the maximum local minimum (I think).

If that's indeed the question, then it's solvable. Unfortunately, I'm super busy today.

Best,

-Slashy

[Spricigo]

10 hours ago, Tig said:

 For instance, each year there is some optimal minimum delta v required to go from low orbit Kerbin to low orbit Eeloo.

The part in bold stablish a time limit. Mind you, is an arbitrary limit (what is special about a year? Or any  other period?) We may calculate thing considering that limit but is important to notice this is a particular case our conclusion may not be valid outside the limits of this particular case.

Two questions:

• What need to happen within a year? (departure, arrival, return)
• from what moment we start to count this year?

Depending on our actual limit, you actual required delta V will be either off a Hohmann transfer or the closest thing to it we can do*.

Calculating the closest thing take a good deal of maths, more than I'm willing to do myself instead of recurring to the transfer planer (exploring diverse limits for earliest/lasted departure and flight time) or something like that. I'd let for each one do the same or if someone more skilled and  not so busy to present a delve in the maths.

 

 

*for the sake of completeness, if the condition are right it may be the third option: a bi-elliptical transfer, but that minimum involves a few infinities. Also it takes much more time than a Hohmann transfer. Similarly we may benefit from gravity assists, what requires a favourable alignment of planets.

[Nicias]

I think @Tig is suggesting that the  "every year" is "every synodic period." (Which is about a year for targets with very long orbits)

Ignoring inclination,  the question is then:  Given two orbits, with given SMA and eccentricities (and AoP). If you departed your original planet at any random transfer window, what would be the maximum DV you would have to spend on the transfer (just counting hyperbolic excess, which is what I think he means by "from the edge of the SOI").  Equivalently. If you left the original planet's orbit (around the sun) at any point in the orbit using a Hohmann transfer to the target planet's orbit (but not necessarily hitting the target planet) what is the most DV you would need?

In other words, there are good transfer windows and bad transfer windows, how bad are the bad ones?

[Spricigo]

4 hours ago, Nicias said:

In other words, there are good transfer windows and bad transfer windows, how bad are the bad ones?

imagine you are in a city and you want to go from place A to place B

Good window. . You open a GPS and follow the best route wich happens to be a straight line.

Bad window. A car just jump in front of you, 4 armed guys take you and throw in the  trunk, and drive away to the to a different state. Federal agents take months to discover where you are and set you free, and only then you arrive at place B after being held in several different places.

 

Change distance covered for deltaV required and that is what it appear for transfer between planets.

 

 

 

 

Edited January 31, 2018 by Spricigo
Do'h missed the point that you didn't need to hit the target planet. (Rockrt science? Pfff!!! Reading is hard.)

[GoSlash27]

I ran the numbers on the worst possible Kerbin->Eeloo Hohmann transfer. I get 4,528.6 m/sec DV. This is assuming that Eeloo is at Pe and the maximum inclination error needs to be corrected at 6.15°. If you run the launch window planner out to an infinite number of years, you should never find a local minimum that exceeds this value.

The Vorb and Vesc never vary about the parents. The changes are all in the Vxs and plane change DV.

The worst possible transfers occur when (perversely) the difference between the planets' altitudes are at a minimum. This is because while it takes a little more DV on the burn out to Eeloo at Ap, it takes a *lot* more DV on the retroburn at Pe because the planet is moving faster at the time. The additional velocity at midcourse adds to the total a bit, but overcoming the velocity difference is the big killer.

The numbers...

Eeloo Ap 1.135E+011m V@Ap 2766.8 VAp 449.5 slower than nom at that altitude
      Pe 6.669E+010m V@Pe 4711.0 VPe 514.1 faster than nom at that altitude
6.15° inclination.

Vxs to Ap 3126.8, save 449.5 on retroburn. V@Midcourse= 6954.0 DV 746.1
Vxs from Ap 1279.2

Vxs to Pe 2684.8, additional 514.1 on retroburn. V@midcourse= 7210.6 DV 773.6
Vxs from Pe 2268.2

Kerbin Vorb 2295.9 Vesc 3246.9
Eeloo Vorb 568.8 Vesc 804.4

Kerbin->Eeloo at Pe sqrt(3246.9^2+2684.8^2)-2295.9= 1,917.2
Eeloo ->Kerbin at Pe sqrt(804.4^2+2268.2^2)-568.8 = 1,837.8
DV plane change at Pe                             =   773.6
                                                  _________
                                                    4,528.6 m/sec

Best,
-Slashy

 

Edited January 31, 2018 by GoSlash27

[Tig]

6 hours ago, Nicias said:

I think @Tig is suggesting that the  "every year" is "every synodic period." (Which is about a year for targets with very long orbits)

Indeed I did.  I tried to write it in a way that a newer player might understand.  The irony of quite a bit of rocket/orbital science is that if you understand how to properly state a question with all proper terminology then you're likely capable of answering the question yourself.

6 hours ago, Nicias said:

Ignoring inclination,  the question is then:

But see I find that inclination is the hardest part for newer players to deal with.  Differences in inclination can greatly increase the needed delta v budget, and is more difficult/unexpected for beginners.  Moho probably is the best example.

2 hours ago, GoSlash27 said:

I ran the numbers on the worst possible Kerbin->Eeloo Hohmann transfer. I get 4,528.6 m/sec DV. This is assuming that Eeloo is at Pe and the maximum inclination error needs to be corrected at 6.15°.

 

This was my intuitive sense, I wasn't sure how to prove it...  I mean, that the worst time to transfer would typically occur at the greatest inclination difference.  Not to say that eccentric orbits don't change dv (your PE vs AP discussion demonstrates that) but just a few degrees of inclination can substantially increase dv. 

[Nicias]

The way to deal with inclination changes is to roll the into one of your burns. Then you get an Oberth boost, but more importantly a Pythagorean savings. For instance, doing a 700 m/s inclination change on top of a 2000 m/s ejection burn only costs: Sqrt(2000^2+700^2)=2119 m/s. 

 

@GoSlash27 I was looking at doing the math for a Hohmann transfer ellipse to ellipse, and found it to be rough. Even just doing ellipse to circle was rough. I couldn't figure out how to find the optimal burn to go from a given point on a given elliptical orbit to an new elliptical orbit with a given periapsis. Is there an easy way? Is it a safe assumption that the burn should be entirely retrograde?

[GoSlash27]

2 hours ago, Nicias said:

The way to deal with inclination changes is to roll the into one of your burns. Then you get an Oberth boost, but more importantly a Pythagorean savings. For instance, doing a 700 m/s inclination change on top of a 2000 m/s ejection burn only costs: Sqrt(2000^2+700^2)=2119 m/s. 

Nicias,
 Unfortunately, it's unlikely that the interplanetary transfer window will happen to coincide with an ascending or descending node. When it does, it's not necessary to perform any inclination change at all; you'll still get there. For simple transfers where there's no specific window, then yeah... roll the inclination change into one of the burns; whichever burn is at the Ap of the transfer.

2 hours ago, Nicias said:

I was looking at doing the math for a Hohmann transfer ellipse to ellipse, and found it to be rough. Even just doing ellipse to circle was rough. I couldn't figure out how to find the optimal burn to go from a given point on a given elliptical orbit to an new elliptical orbit with a given periapsis. Is there an easy way? Is it a safe assumption that the burn should be entirely retrograde?

Yeah, the answers to that are as fuzzy as the math. Your best bet is if the orbits are coplanar, do the transfer where they are farthest apart; physically largest transfer orbit. If they have a substantial inclination difference, do the transfer at the ascending or descending node, whichever allows you to fold in the plane change at the higher transfer altitude. If there's a difference in the argument of periapsis between the two orbits, you will have to incorporate some radial in or out as well. Again, at the higher transfer altitude.

 There's a lot of gray area in the math, but sticking to these rules of thumb helps to keep you under the DV budget.

Best,
-Slashy

[Kryxal]

You can actually manage some savings by NOT doing all your normal burn as far out as possible ... incorporate it into ALL your burns since it's so cheap for lower angles.  It can make things more tricky, though.

[OhioBob]

22 hours ago, Nicias said:

The way to deal with inclination changes is to roll the into one of your burns. Then you get an Oberth boost, but more importantly a Pythagorean savings. For instance, doing a 700 m/s inclination change on top of a 2000 m/s ejection burn only costs: Sqrt(2000^2+700^2)=2119 m/s.

You definitely want to try to combine a plane change with an altitude change when ever possible to take advantage of the Pythagorean savings.  However, depending on what it is you are trying to do, figuring out the best way to take advantage of it can still get a little complicated.  For instance, let's consider the real life situation of transferring a satellite from low earth orbit to geostationary orbit.  When a satellite is launched from Cape Canaveral it's going to end up in a low orbit with an inclination of about 29 degrees.  So where do we make the plane change it get the satellite into a zero inclination orbit?  It is most economical to perform the majority of the plane change in combination with the circularization burn at apogee of the transfer orbit.  However, because of the Pythagorean principal, it's actually possible to get a couple degrees of plane change for virtually free by combining it with the low orbit burn that places the satellite into its transfer orbit.  So the plane change is done in stages, with about 2 degree being done in the first burn, and about 27 degrees in the second burn.  This results in the lowest total delta-v.
 

[Nicias]

I am trying to run the numbers on this, and I think I almost have it. I just need to add in the inclination change. I have a couple of questions. 

1) I'm trying to use the formula for an inclination change given at: https://en.wikipedia.org/wiki/Orbital_inclination_change

My question is in that expression, if the inclination change is happening at one of the nodes, then f is 180 - w or  -w right?  w measures the angle between the semi major axis and the line of nodes, and f measures the angle between the current position and the sma.  So if we are at one of the nodes, then f is the measure to the next node from the sma, so  since the nodes are 180 degrees apart, its either -w (if we are changing at the ascending node,) or 180 - w  (if we are changing at the desending node), 

That would make w+f  180 or 0  right?  

2) I am parametrizing the transfer orbits by true anomaly (f) and eccentricity of the transfer orbit.  To use the formula for inclination change I need the argument of periapsis for the transfer orbit. I know my radius from the sun at departure (r),  I can work out the new aop (w) using: 

r = a * (1-e^2)/(1-e Cos[f-w])

Solving this for w gives:

w=  f +/-  ArcCos[ (a (1-e^2)-r)/  e r  ] 

With +/- depending on if I my current radial velocity is outward or inward at this time. 

Does that seem correct?

With that done, added in, I think I could definitely work out the problem for any planet to/form kerbin. If one end is not circular, I don't think I could get it to work.

 

 

[GoSlash27]

Nicias,

 I'm afraid I can't even begin to decode what you just asked.

An inclination change is simply 2sin(theta/2)* V at the instant you make it; exactly the same as a purely Newtonian vector change in 2d space. This probably doesn't answer your question, but I hope it at least helps.

Best,
-Slashy

[Nicias]

Slashy,

Thanks for giving it a look :). That change is correct, the difficulty is in finding theta. Theta is the angle between the vectors. It is only the angle between the orbits if the velocity is purely tangential (so at apoapsis or periapsis). Take the limit of two orbits that 90degrees inclined to each other, but highly elliptical. If they have 1000 m/s radial and 1m/s tangential, then the angle between the two velocities is only about 0.08 degrees, rather than the 90 degrees.  So you need to find that angle, it isn't just the relative inclination. Or you have to find the tangential velocity, which is easier, because it is just given by  h/r (where h is the angular momentum of the orbit) but then you need to find the radius when you do the burn. Which means you need to find out where on the transfer ellipse the AN/DN is, which means you need to find the AoP of the transfer ellipse, which leads to my question about AoP.

Basically, I am asking how you calculated velocity at the inclination change?

-N

 

Edited February 2, 2018 by Nicias
(added clarification question)

[GoSlash27]

46 minutes ago, Nicias said:

Basically, I am asking how you calculated velocity at the inclination change?

Nicias,

I calculated the midcourse burn assuming it's literally midcourse.

(Vap+Vpe)sin(theta/2)

 Thinking about it further... I may have overestimated the velocity. I'll have to take another look at it.

Best,

-Slashy

Edited February 2, 2018 by GoSlash27