Oberth effect

[Teutooni]

As I understand it, to leave Kerbin SOI, the minimum you need is the Hohmann transfer orbit to Kerbin SOI. Plugging the numbers into the transfer equation: sqrt(u/r1)(sqrt(2*r2/(r1+r2))-1), where u = G*M, r1 is initial circular orbit (670000 m) and r2 is the edge of SOI (84 159 286 m), I get 937.854 m/s. Which seems about right according to my experience. Add the orbital velocity (I was lazy and used browser calculator: http://files.arklyffe.com/orbcalc.html) of 2295.5 m/s and the burnout is 3233.4 m/s. The actual escape velocity from Kerbin, using Ve = sqrt(2*u/r), is 3245.9 m/s from 70km circular orbit. There is a difference of 12.5 m/s. Which seems to confirm what OhioBob just posted while I was writing this.

[OhioBob]

As I understand it, to leave Kerbin SOI, the minimum you need is the Hohmann transfer orbit to Kerbin SOI. Plugging the numbers into the transfer equation: sqrt(u/r1)(sqrt(2*r2/(r1+r2))-1), where u = G*M, r1 is initial circular orbit (670000 m) and r2 is the edge of SOI (84 159 286 m), I get 937.854 m/s. Which seems about right according to my experience. Add the orbital velocity (I was lazy and used browser calculator: http://files.arklyffe.com/orbcalc.html) of 2295.5 m/s and the burnout is 3233.4 m/s. The actual escape velocity from Kerbin, using Ve = sqrt(2*u/r), is 3245.9 m/s from 70km circular orbit. There is a difference of 12.5 m/s. Which seems to confirm what OhioBob just posted while I was writing this.

Yep, that's another good way of examining the problem. Nice confirmation.

[Volix]

Yes, those numbers agree much better . The calculation I did was something of a shot in the dark. It probably still represents something, just maybe not something useful. All around fun discussion though, I hadn't taken the Oberth effect past "burning at higher speeds changes energy more for the same dV".

[match]

I haven't taken the time to look at it in depth yet, but for anyone wanting to see more orbital mechanics math, here's a page that seems to go into a lot of detail:

http://www.braeunig.us/space/orbmech.htm

This should be an entry in "You know you play too much KSP when..." thread. I recognized that website to be OhioBob's the moment I saw it.

One point from the original post that has always struck me as one of the more glossed over aspects of the Oberth Effect is the assumption of infinite thrust at a singular point. I recollect that the energy savings is relative to the current velocity against the average velocity of the orbit (that is not precise, but somewhere along those lines).

Not only does the greater velocity at PE provide the effect, but application of the thrust in the shortest time period also contributes to the savings via OE. That is, the longer the burn, the more distant the vessel travels from the optimal point to burn, therefore, less optimal.

Is this even worth mentioning/examining? My initial feeling is that less eccentric orbits (such as the 350 km x 75 km) will achieve far less benefit from the effect with a low TWR vehicle. A 5 minute burn would be roughly 12% of its original period.

My apologies, as this veers away from the current theory/application endeavor, just felt my brain trying to crinkle around this while reading through this thread.

[OhioBob]

Volix said:

Yes, those numbers agree much better . The calculation I did was something of a shot in the dark. It probably still represents something, just maybe not something useful.

Well, the equation you derived,

v = (2*G*M/x)^1/2

is the equation for escape velocity. So your number of 289.57 m/s is the escape velocity from an orbit with a radius equal to the SOI. Of course, if we're in an orbit out near the SOI, then we just need a small nudge to escape, not 289.57 m/s. However, the computed amount doesn't represent a real savings unless we start out in an orbit that's already out near the SOI. If we're ejecting from a low orbit, our savings are far less because it takes only a small amount of addition velocity when close to Kerbin to have a large amount left over when we reach the SOI (that's the Oberth effect).

For example, Teutooni computed that, from a 70 km orbit, it takes 938 m/s to just reach the SOI. If we were to cross the SOI at that point we would have 0 residual velocity. We can now use my equation,

V_soi² = 2μ * (1/R_soi - 1/R_bo) + V_bo²

to compute how much Δv we need to provide at 70 km to have 289.57 m/s left over after crossing the SOI.

V_bo = (289.57² - 2*3.5316E+12*(1/84159286 - 1/670000))^0.5 = 3247 m/s

Δv = V_bo - V_orb = 3247 - 2296 = 951 m/s

Therefore it takes only 951 - 938 = 13 m/s additional Δv.

The number you computed is useful, it's just not what you thought it was.
 

Edited April 9, 2016 by OhioBob

[Teutooni]

Incidentally, using just maneuver nodes, it takes exactly 951 m/s to reach Kerbin escape from 70.1 km orbit. As you can see in this picture, 950.9 m/s is still "inside the SOI", with apparent apoapsis well over Jool's orbit. I think the node system uses infinitely large SOI, i.e. real escape velocity. Doing the actual burn instead of fiddling with nodes gives numbers close to 938 m/s and 84Mm SOI radius.

[Birdco_Space]

Ohiobob, 2 Questions I have on this topic.

1) has anyone analyzed the dV differences with getting to different parking orbits? I am speaking of straight burns to the desired apoapsis. I've found that if I do a good gravity turn (nearly horizontal at 60km), there isn't a huge dV difference between getting to different apoapsis. I think where the extra cost comes is in the circulation burn. Is 75km truly the best altitude to park at while preparing for a departure? Maybe timing your launch to have the resulting periapsis of a highly elliptical orbit you launch from directly generally where you want the departure burn, a short burn at apoapsis to get the periapsis out of the atmosphere, then a final burn out to the required hyperbolic excess velocity at periapsis results in the most efficient escape.

2) How does thrust to weigh ratio affect the optimal altitude calculations? is there a way to factor cosine losses into these charts to figure the most dV efficient exit burn for high excess velocity situations? I've found that an Ion probe cannot efficiently eject itself from Kerbin SOI to get to Moho with a 75km periapsis. Very high dV (thus very low TWR) nuclear rockets have similar issues.

[mikegarrison]

2) How does thrust to weigh ratio affect the optimal altitude calculations? is there a way to factor cosine losses into these charts to figure the most dV efficient exit burn for high excess velocity situations? I've found that an Ion probe cannot efficiently eject itself from Kerbin SOI to get to Moho with a 75km periapsis. Very high dV (thus very low TWR) nuclear rockets have similar issues.

The issue there is that the assumptions behind the math are that the dV is added instantaneously. However, it actually takes some amount of time to add the dV. This is an inefficiency because you are no longer adding the dV at the PE. It's even more complicated because your orbital speed as you are adding the last of your dV is quite a bit higher than your orbital speed as you are adding the first of your dV, so in Oberth terms the end of your burn should be more efficient than the beginning of your burn.

Anyway, personally I prefer to use a high efficiency but low thrust engine and am therefore not terribly concerned about a few extra 100 m/s if I don't choose the theoretical optimum. I am more worried about things like steering losses, so for those looong burns I prefer to do them in high orbits.

[OhioBob]

Birdco_Space said:

1) has anyone analyzed the dV differences with getting to different parking orbits? I am speaking of straight burns to the desired apoapsis. I've found that if I do a good gravity turn (nearly horizontal at 60km), there isn't a huge dV difference between getting to different apoapsis. I think where the extra cost comes is in the circulation burn. Is 75km truly the best altitude to park at while preparing for a departure? Maybe timing your launch to have the resulting periapsis of a highly elliptical orbit you launch from directly generally where you want the departure burn, a short burn at apoapsis to get the periapsis out of the atmosphere, then a final burn out to the required hyperbolic excess velocity at periapsis results in the most efficient escape.

We had a recent discussion in another thread in which we discussed the economics of reaching high Kerbin orbits, i.e. a few hundred kilometers. A couple of us analyzed the problem mathematical and found that Δv required to launch directly into a high orbit is almost identical to the Δv required to first insert into a minimum orbit, and then perform a Hohmann transfer to the higher orbit. In other words, we can go directly from the ground to a 250 km orbit for the same Δv as it would take to go from the ground to a 70 km orbit, and then from 70 km to 250 km.

We can, therefore, assume that the Δv to reach 70 km is some fixed standard amount. The cost to reach higher orbits can then be evaluated strictly on the basis of the Hohmann transfer Δv. To determine which orbit provides the best overall economics, we simply have to add the Hohmann transfer Δv to the ejection Δv and see which orbit is gives the lowest total Δv.

Below is an example based on V_∞ = 2800 m/s, i.e. a trip to Jool. We see that the lowest ejection Δv occurs around 300 km, which is what we determined earlier in this thread. However, the lowest total Δv occurs at 70 km, with the total Δv constantly increasing with increasing altitude. If I were to produce a similar table for other values of V_∞, we would see the same trend. When launching from the ground, it is always better to seek out the lowest possible orbit.

Altitude, Z (km)	Δv from 70 km to Z (m/s)	Ejection Δv (m/s)	Total Δv (m/s)
70	0.0	1991.6	1991.6
80	16.9	1990.4	2007.3
90	33.5	1989.3	2022.8
100	49.7	1988.3	2038.0
120	81.1	1986.5	2067.6
140	111.2	1984.9	2096.1
160	140.1	1983.6	2123.7
180	167.8	1982.6	2150.4
200	194.4	1981.7	2176.1
250	256.6	1980.3	2237.0
300	313.3	1979.9	2293.2
350	365.0	1980.2	2345.3
400	412.5	1981.2	2393.7
450	456.2	1982.7	2438.9
500	496.5	1984.6	2481.1

 

Birdco_Space said:

2) How does thrust to weigh ratio affect the optimal altitude calculations? is there a way to factor cosine losses into these charts to figure the most dV efficient exit burn for high excess velocity situations? I've found that an Ion probe cannot efficiently eject itself from Kerbin SOI to get to Moho with a 75km periapsis. Very high dV (thus very low TWR) nuclear rockets have similar issues.

That's an issue I haven't studied in detail. I have no reply at this time.

(edited to add)

OK, I've thought about this and I've come to the conclusion that the problem of minimizing losses due to low TWR should be studied independently of what we've been discussing in this thread. Although the basic question is the same, i.e. are the potential Δv savings more than the Δv needed to raise the orbit, linking to two problems together needlessly complicates things. You can see from the table above that, for high V_∞, the ejection Δv savings that comes with increasing altitude are so small that it's really not worth considering when studying the TWR problem. To properly study the TWR problem, somebody needs to figure out how much Δv is saved by moving to higher orbits and compare that number to the values in column 2 above, i.e. "Δv from 70 km to Z." If you want to, you can add the TWR savings and the ejection savings together, but I think they should be considered separately.

 

Edited April 9, 2016 by OhioBob

[NecroBones]

The issue there is that the assumptions behind the math are that the dV is added instantaneously. However, it actually takes some amount of time to add the dV. This is an inefficiency because you are no longer adding the dV at the PE.

Just for clarification (because someone tried to school me on the same thing on the KSP subreddit a while back): It's actually that you're spending time burning away from the maneuver node, or away from your velocity vector, not necessarily the PE.

[mikegarrison]

Just for clarification (because someone tried to school me on the same thing on the KSP subreddit a while back): It's actually that you're spending time burning away from the maneuver node, or away from your velocity vector, not necessarily the PE.

No, recall that we are talking about Oberth here. The basic fundamental of Oberth is that it is more efficient to add velocity to a fast rocket than a slow one. So the drop in Oberth efficiency is, in fact, due to you not being at PE (where your orbital speed is fastest). However, that is complicated by the issue that as you steadily burn your rocket you are constantly adding velocity. So for Oberth reasons, your dV becomes more and more effective over the course of your long burn.

This is an effect that is in addition to the so-called "steering losses" involved in burning off of prograde.

[OhioBob]

OhioBob said:

Birdco_Space said:

2) How does thrust to weigh ratio affect the optimal altitude calculations? is there a way to factor cosine losses into these charts to figure the most dV efficient exit burn for high excess velocity situations? I've found that an Ion probe cannot efficiently eject itself from Kerbin SOI to get to Moho with a 75km periapsis. Very high dV (thus very low TWR) nuclear rockets have similar issues.

That's an issue I haven't studied in detail. I have no reply at this time.

(edited to add)

OK, I've thought about this and I've come to the conclusion that the problem of minimizing losses due to low TWR should be studied independently of what we've been discussing in this thread. Although the basic question is the same, i.e. are the potential Δv savings more than the Δv needed to raise the orbit, linking to two problems together needlessly complicates things. You can see from the table above that, for high V_∞, the ejection Δv savings that comes with increasing altitude are so small that it's really not worth considering when studying the TWR problem. To properly study the TWR problem, somebody needs to figure out how much Δv is saved by moving to higher orbits and compare that number to the values in column 2 above, i.e. "Δv from 70 km to Z." If you want to, you can add the TWR savings and the ejection savings together, but I think they should be considered separately. .

I've studied this further and have some numbers to share. Despite my complaint that it would be too hard to study both effects at the same time, it actually turned out to be quite easy. What I did was to simulate ejection burns from different orbits at different TWR. I ran the simulation until the hyperbolic excess velocity reached 2800 m/s, at which point I cut the engine and figured out how much Δv was used. Throughout the burn I kept the rocket pointed prograde, which may not be exactly what is done in the game, but it was easy to simulate. I assumed a specific impulse of 345 s.

Earlier we saw that the optimum ejection altitude, independent of TWR losses, was about 300 km. When we now take into consideration that TWR losses decrease with increasing altitude, we push the optimum altitude a bit higher. To about 350 km for moderately high TWR, and to much higher for very low TWR. However, in all cases the savings still aren't enough to offset the Δv needed to get to the higher orbit. If we're launching from the ground, it's still going to require fewer Δv to park in a low orbit and absorb the other losses that might be incurred.

Ejection ΔV, for V∞=2800 m/s
Alt.	Acceleration at start (g)
(km)	1.5	1	0.5	0.2
75	1999.0	2008.6	2054.3	2247.2
100	1995.5	2004.2	2045.9	2225.7
150	1990.2	1997.4	2032.3	2189.4
200	1986.7	1992.7	2022.3	2160.3
250	1984.5	1989.6	2015.0	2136.8
300	1983.5	1987.8	2009.7	2117.9
350	1983.3	1987.1	2006.1	2102.6
400	1983.9	1987.2	2003.8	2090.2
450	1985.0	1987.9	2002.6	2080.3
500	1986.6	1989.2	2002.2	2072.3
550	---	---	2002.5	2066.0
600	---	---	2003.3	2061.1
650	---	---	2004.6	2057.3
700	---	---	---	2054.5
750	---	---	---	2052.5
800	---	---	---	2051.2
850	---	---	---	2050.5
900	---	---	---	2050.3
950	---	---	---	2050.5
1000	---	---	---	2051.1

Edited May 24, 2016 by OhioBob

[GoSlash27]

I’ve studied this further and have some numbers to share. Despite my complaint that it would be too hard to study both effects at the same time, it actually turned out to be quite easy. What I did was to simulate ejection burns from different orbits at different TWR. I ran the simulation until the hyperbolic excess velocity reached 2800 m/s, at which point I cut the engine and figured out much ÃŽâ€v was used. Throughout the burn I kept the rocket pointed prograde, which may not be exactly what is done in the game, but it was easy to simulate. I assumed a specific impulse of 345 s.

Earlier we saw that the optimum ejection altitude, independent of TWR losses, was about 300 km. When we now take into consideration that TWR losses decrease with increasing altitude, we push the optimum altitude a bit higher. To about 350 km for moderately high TWR, and to much higher for very low TWR. However, in all cases the savings still aren’t enough to offset the ÃŽâ€v needed to get to the higher orbit. If we’re launching from the ground, it’s still going to require fewer ÃŽâ€v to park in a low orbit and absorb the other losses that might be incurred.

[TABLE=width: 320]

[TR]

[TD=colspan: 5, align: center]Ejection ÃŽâ€V, for V_{∞}=2800 m/s[/TD]

[/TR]

[TR]

[TD=align: center]Alt.[/TD]

[TD=colspan: 4, align: center]Acceleration at start (g)[/TD]

[/TR]

[TR]

[TD=align: center](km)[/TD]

[TD=align: center]1.5[/TD]

[TD=align: center]1[/TD]

[TD=align: center]0.5[/TD]

[TD=align: center]0.2[/TD]

[/TR]

[TR]

[TD=align: center]75[/TD]

[TD=align: center]1999.0[/TD]

[TD=align: center]2008.6[/TD]

[TD=align: center]2054.3[/TD]

[TD=align: center]2247.2[/TD]

[/TR]

[TR]

[TD=align: center]100[/TD]

[TD=align: center]1995.5[/TD]

[TD=align: center]2004.2[/TD]

[TD=align: center]2045.9[/TD]

[TD=align: center]2225.7[/TD]

[/TR]

[TR]

[TD=align: center]150[/TD]

[TD=align: center]1990.2[/TD]

[TD=align: center]1997.4[/TD]

[TD=align: center]2032.3[/TD]

[TD=align: center]2189.4[/TD]

[/TR]

[TR]

[TD=align: center]200[/TD]

[TD=align: center]1986.7[/TD]

[TD=align: center]1992.7[/TD]

[TD=align: center]2022.3[/TD]

[TD=align: center]2160.3[/TD]

[/TR]

[TR]

[TD=align: center]250[/TD]

[TD=align: center]1984.5[/TD]

[TD=align: center]1989.6[/TD]

[TD=align: center]2015.0[/TD]

[TD=align: center]2136.8[/TD]

[/TR]

[TR]

[TD=align: center]300[/TD]

[TD=align: center]1983.5[/TD]

[TD=align: center]1987.8[/TD]

[TD=align: center]2009.7[/TD]

[TD=align: center]2117.9[/TD]

[/TR]

[TR]

[TD=align: center]350[/TD]

[TD=align: center]1983.3[/TD]

[TD=align: center]1987.1[/TD]

[TD=align: center]2006.1[/TD]

[TD=align: center]2102.6[/TD]

[/TR]

[TR]

[TD=align: center]400[/TD]

[TD=align: center]1983.9[/TD]

[TD=align: center]1987.2[/TD]

[TD=align: center]2003.8[/TD]

[TD=align: center]2090.2[/TD]

[/TR]

[TR]

[TD=align: center]450[/TD]

[TD=align: center]1985.0[/TD]

[TD=align: center]1987.9[/TD]

[TD=align: center]2002.6[/TD]

[TD=align: center]2080.3[/TD]

[/TR]

[TR]

[TD=align: center]500[/TD]

[TD=align: center]1986.6[/TD]

[TD=align: center]1989.2[/TD]

[TD=align: center]2002.2[/TD]

[TD=align: center]2072.3[/TD]

[/TR]

[TR]

[TD=align: center]550[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2002.5[/TD]

[TD=align: center]2066.0[/TD]

[/TR]

[TR]

[TD=align: center]600[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2003.3[/TD]

[TD=align: center]2061.1[/TD]

[/TR]

[TR]

[TD=align: center]650[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2004.6[/TD]

[TD=align: center]2057.3[/TD]

[/TR]

[TR]

[TD=align: center]700[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2054.5[/TD]

[/TR]

[TR]

[TD=align: center]750[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2052.5[/TD]

[/TR]

[TR]

[TD=align: center]800[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2051.2[/TD]

[/TR]

[TR]

[TD=align: center]850[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2050.5[/TD]

[/TR]

[TR]

[TD=align: center]900[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2050.3[/TD]

[/TR]

[TR]

[TD=align: center]950[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2050.5[/TD]

[/TR]

[TR]

[TD=align: center]1000[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]---[/TD]

[TD=align: center]2051.1[/TD]

[/TR]

[/TABLE]

OhioBob,

This is why I design my vacuum missions with a minimum acceleration of .5G. I've noticed that the DV budget grows exponentially below that.

Best,

-Slashy

[Red Iron Crown]

I'm betting OhioBob is basing the TWR losses from doing the ejection as a single burn, in practice I have found that splitting the burn up can reduce those losses significantly. I wonder how those numbers change if the starting orbit has an Ap near the SoI edge and Pe altitude as a variable?

(Not meant to take away from the excellent analysis so far, the ability to crunch these numbers quickly is enviable and I'm learning a lot.)

[GoSlash27]

I'm betting OhioBob is basing the TWR losses from doing the ejection as a single burn, in practice I have found that splitting the burn up can reduce those losses significantly. I wonder how those numbers change if the starting orbit has an Ap near the SoI edge and Pe altitude as a variable?

(Not meant to take away from the excellent analysis so far, the ability to crunch these numbers quickly is enviable and I'm learning a lot.)

RIC,

I'm sure this is the case as well. Executing multiple Apoapsis kicks can maintain high efficiency with ridiculously low t/w ratios, but working out timing for transfer windows gets ugly in a hurry. Errors in timing the transfer add a whole other source of inefficiency.

I'm afraid it all gets subjective at this point

Best,

-Slashy

[NecroBones]

No, recall that we are talking about Oberth here. The basic fundamental of Oberth is that it is more efficient to add velocity to a fast rocket than a slow one. So the drop in Oberth efficiency is, in fact, due to you not being at PE (where your orbital speed is fastest). However, that is complicated by the issue that as you steadily burn your rocket you are constantly adding velocity. So for Oberth reasons, your dV becomes more and more effective over the course of your long burn.

This is an effect that is in addition to the so-called "steering losses" involved in burning off of prograde.

Fair enough, since we are talking about Oberth specifically.

It's funny. I was making the argument you are, and a Boeing employee started in to me about steering losses and putting too much emphasis on Periapsis. I give up. lol

[FleshJeb]

I think this is my favorite KSP thread so far.

I have a couple of points to add that haven't been mentioned yet. (And please correct any misconceptions I have.)

1. We can do a very similar analysis for returns from moons to parent body Pe's.

2. None of the analysis upthread applies to or accounts for Normal/Anti components of your ejection burn.

2a. Plane change burns do not get ANY benefit from Oberth.

2b. Plane change burns are least expensive at the lowest orbital velocity.

2c. The scenario that would prove this best/be a worst case-scenario, is a direct ejection to Moho.

3. The analysis upthread should still be valid for bi-elliptic transfers.

3a. The only Kerbin transfer that benefits from bi-elliptic is Eeloo, and this should also be in OhioBob's chart.

4. Jeb's Transfer: Gravity slingshotting a giant, overpowered, phallic ship named the Llewellyn Dowd off of Eve is STILL the BEST method

[OhioBob]

GoSlash27 said:

This is why I design my vacuum missions with a minimum acceleration of .5G.

Same here. Based on a similar analysis I had done earlier, 0.5g seemed to be the border between minimal losses and significantly mounting losses.

 

Red Iron Crown said:

I'm betting OhioBob is basing the TWR losses from doing the ejection as a single burn.

Absolutely. The way the question was asked, I assumed the inquiry was about a single burn.

 

Red Iron Crown said:

I wonder how those numbers change if the starting orbit has an Ap near the SoI edge and Pe altitude as a variable?

I'm not sure I understand. Are you wondering if there is an optimum Pe at which you should finish off your ejection burn if you're in an orbit with an Ap near the SoI? If that's what you're asking, then I wondered something similar a couple days ago and played around with some numbers. I think the answer is that the Pe should be as low as possible, i.e. just above the atmosphere. I'll look at it again to make sure. If I've misunderstood what you are asking, then please rephrase the question.

 

GoSlash27 said:

...but working out timing for transfer windows gets ugly in a hurry. Errors in timing the transfer add a whole other source of inefficiency.

I rarely have TWR so low than I can't do it in one burn. However, on the couple occasions when I did perform multiple burns, I performed the first burn exactly one day before my planned launch window and put the spacecraft into an orbit with a 6-hour period. This requires a Δv of about 790 m/s and brings the spacecraft right back to Pe on schedule to complete the ejection the next day.

- - - Updated - - -

FleshJeb said:

2b. Plane change burns are least expensive at the lowest orbital velocity.

This is definitely true when performing a stand-alone plane change maneuver. However, in some instances the cheapest way to perform a plane change is in combination with an altitude change. In the case of ejection and insertion burns, a plane change of up to a few degrees can be made for a trivial amount of additional Δv. For instance, if our ejection is 1000 m/s prograde and 100 m/s normal, the total burn is (1000²+100²)^0.5 = 1005 m/s. We get a 100 m/s plane change done for just 5 m/s.
 

Edited April 9, 2016 by OhioBob

[Red Iron Crown]

I'm not sure I understand. Are you wondering if there is an optimum Pe at which you should finish off your ejection burn if you're in an orbit with an Ap near the SoI? If that's what you're asking, then I wondered something similar a couple days ago and played around with some numbers. I think the answer is that the Pe should be as low as possible, i.e. just above the atmosphere. I'll look at it again to make sure. If I've misunderstood what you are asking, then please rephrase the question.

Bolded section is what I meant. If using an interplanetary ship being refueled from Minmus, I dock a tanker to it in LKO, burn Ap up to Minmus altitude, then refuel, undock the tanker and complete the ejection. I've been keeping the LKO altitude as low as practical to maximize Oberth effect, but this thread is making me think that a higher altitude might result in a cheaper burn after undocking the tanker.

[OhioBob]

Red Iron Crown said:

Bolded section is what I meant. If using an interplanetary ship being refueled from Minmus, I dock a tanker to it in LKO, burn Ap up to Minmus altitude, then refuel, undock the tanker and complete the ejection. I've been keeping the LKO altitude as low as practical to maximize Oberth effect, but this thread is making me think that a higher altitude might result in a cheaper burn after undocking the tanker.

I assume that once you boost your Ap to Minmus, you remain in that elliptical orbit until you complete the ejection. That is, you coast up to Ap, undock the tanker, coast back down to Pe, and perform the final ejection. If that's correct, then having your Pe as close to Kerbin as possible results in the lowest Δv for your final burn.

A couple days ago I was discussing two-burn scenarios with Slashy. In that discussion the assumption was that we were starting out in a high circular orbit. From that orbit a burn would be made to lower the Pe, and then the ejection burn would be made at Pe. What I found was that there was an altitude below which the savings gained from ejecting at a low Pe was less than the Δv it took to lower the Pe. In that case it was most economical to eject directly from the initial circular orbit. If, on the other hand, our initial orbit is above the critical altitude, then the gain we get from burning at the low Pe is greater than the Δv it takes to lower the Pe. In that case we should lower the Pe; however, when we lower it, we should lower it as far as possible to take greatest advantage of the Oberth effect.
 

Edited April 9, 2016 by OhioBob

[mikegarrison]

Fair enough, since we are talking about Oberth specifically.

It's funny. I was making the argument you are, and a Boeing employee started in to me about steering losses and putting too much emphasis on Periapsis. I give up. lol

Oh, I think the steering losses are the bigger effect. It's just that this thread was really only about Oberth.

[Red Iron Crown]

I assume that once you boost your Ap to Minmus, you remain in that elliptical orbit until you complete the ejection. That is, you coast up to Ap, undock the tanker, coast back down to Pe, and immediately perform the final ejection. If that's correct, then having your Pe as close to Kerbin as possible results in the lowest ÃŽâ€v for your final burn.

Actually I don't complete another orbit, I just undock the tanker and immediately resume the ejection burn. But I think that scenario is mathematically equivalent to the one you describe (aside from the combined burns taking a bit longer).

A couple days ago I was discussing two-burn scenarios with Slashy. In that discussion the assumption was that we were starting out in a high circular orbit. From that orbit a burn would be made to lower the Pe, and then the ejection burn would be made at Pe. What I found was that there was an altitude below which the savings gained from ejecting at a low Pe was less than the ÃŽâ€v it took to lower the Pe. In that case it was most economical to eject directly from the initial circular orbit. If, on the other hand, our initial orbit is above the critical altitude, then the gain we get from burning at the low Pe is greater than the ÃŽâ€v it takes to lower the Pe. In that case we should lower the Pe; however, when we lower it, we should lower it as far as possible to take greatest advantage of the Oberth effect.

Thank you, this reassures me that I haven't been doing it wrong (or at least wrong in this particular way ).

[magnemoe]

OhioBob,

This is why I design my vacuum missions with a minimum acceleration of .5G. I've noticed that the DV budget grows exponentially below that.

Best,

-Slashy

Howerver its hard to get an large ship up to 0.5 g with LV-N.

In my experience the error you get then doing an burn who takes long time is war worse than the Oberth effect losses.

Most of this however can be solved by doing it in two burns or starting from Minmus.

[OhioBob]

magnemoe said:

Howerver its hard to get an large ship up to 0.5 g with LV-N.

In my experience the error you get then doing an burn who takes long time is war worse than the Oberth effect losses.

Most of this however can be solved by doing it in two burns or starting from Minmus.

From a strictly mathematical perspective, the numbers say that is worse to perform the burn from a high orbit. Although the burn losses can be significantly reduced, they aren't reduced enough to offset the Δv required to reach the high orbit. However, from my experience I agree that there are other factors that favor using a high orbit for very long burns. For instance, burn inaccuracies. With short burns I find it easy to hit my intercept with the target planet. However, when I perform a long burn from a low orbit, I'm rarely able to hit the spot I was aiming for because the errors start to mount. I usually end up having to make a second burn to correct my course and get back on target with a good intercept. Targeting errors and course corrections are not factored into the mathematical analysis, but they can be a real problem when playing the game. A large course correction could swing things back in favor of the high orbit in terms of Δv. Ejecting from a high orbit can certainly make execution easier even if the economics are debatable.
 

Edited April 9, 2016 by OhioBob

[Archgeek]

I dock a tanker to it in LKO, burn Ap up to Minmus altitude, then refuel, undock the tanker and complete the ejection.

That's a neat technique, there! I'm kinda surprised you aren't having the tanker act more as a Von Braun tug, though -- gassing up the mission craft, completing all or as much of the transfer burn as it can get away with, then undocking and slowing itself back down to head back to Minmus.