Oberth effect

[xtoro]

If you want to compare starting orbits and their savings, you can't just compare starting points from different altitudes unless you're completely negating whatever method you used to GET to that orbit. You need the entire spent dV from the surface to that orbit with the same ship, same mass, same aerodynamics (if launching from atmospheric planet), same flight profile etc.

Getting to 70km and getting to 300km circular orbits don't cost the same in dV. Also, at 300km, your orbital speed may be slower but the effects of gravity are weaker and thus easier to escape from with less dV.

BUT, this about OE, not ejection burns so sorry to continue the topic derailment...

- - - Updated - - -

OhioBob,

I always send my missions assembled and fueled in orbit.

I'm certain that the most economical (from a DV standpoint) approach is to start at a higher altitude, drop the periapsis, then hit the transfer. But launching direct from higher up? It's never worked out in my practice.

I've bookmarked this page. I'll have to ponder this...

Best,

-Slashy

If you already have a high circular orbit, why would you spend dV to slow yourself down just to pick up speed again?

See my comment on gravity getting weaker the further out you are. As a test, hyperedit your ship out to I think about 68,000 Km circular orbit and see how much dV it takes you to escape Kerbin's SOI. This is of course extremely exaggerated when comparing 68,000 KM to 70KM but it shows there is a difference between X and Y and required dV.

Edited October 6, 2015 by xtoro

[GoSlash27]

If you already have a high circular orbit, why would you spend dV to slow yourself down just to pick up speed again?

xtoro,

Because the small price in DV for the retroburn is more than made up for in velocity when you do the actual ejection burn.

When you're orbiting at high altitude, your energy difference is potential rather than kinetic. Kinetic energy is what gets you places.

Retroburning to lower Pe transforms your potential energy into kinetic energy, allowing you to save DV on the transfer.

As an example, try it both ways from 6.8Mm. The two burn maneuver will cost you a lot less DV in the end.

Trust me, we've been all around this subject with lots of empirical testing to hash it out.

Best,

-Slashy

Edited October 7, 2015 by GoSlash27

[Norcalplanner]

xtoro,

Because the small price in DV for the retroburn is more than made up for in velocity when you do the actual ejection burn.

When you're orbiting at high altitude, your energy difference is potential rather than kinetic. Kinetic energy is what gets you places.

Retroburning to lower Pe transforms your potential energy into kinetic energy, allowing you to save DV on the transfer.

As an example, try it both ways from 6.8Mm. The two burn maneuver will cost you a lot less DV in the end.

Trust me, we've been all around this subject with lots of empirical testing to hash it out.

Best,

-Slashy

Slashy, where would you say the lines cross? Certainly if you're up in orbit around the Mun or Minmus, it makes sense to do what you're talking about. What about something that's just a little higher than LKO, like my favorite fuel depot orbit of 250 km? How about 400 km or 600 km? Is it still worth doing a retro burn first, then burning at Pe from these orbits? My gut says no, but I'm very open to being wrong.

[GoSlash27]

Norcalplanner,

It should always show a DV savings to retroburn to LKO first, no matter what. But the benefit of doing it would diminish with lower starting altitudes. Eventually to the point where it's not saving enough to be worth the trouble. I don't think there's a hard line where it's "worth it", especially when you factor in cosine losses. At some point, the savings are lost in the noise floor and you'd say "Meh. Let's just go".

Best,

-Slashy

Edited October 7, 2015 by GoSlash27

[Superfluous J]

Why restrict raising the periapsis (seemingly for free) just to the Kerbin SOI? What if we start our ship high up in Sun orbit, say right around the orbit of the target planet? I could lower dV to reach said planet to 0 with some luck or planning.

[OhioBob]

Norcalplanner said:

Slashy, where would you say the lines cross? Certainly if you're up in orbit around the Mun or Minmus, it makes sense to do what you're talking about. What about something that's just a little higher than LKO, like my favorite fuel depot orbit of 250 km? How about 400 km or 600 km? Is it still worth doing a retro burn first, then burning at Pe from these orbits? My gut says no, but I'm very open to being wrong.

I've been playing around with the math and I find that the answer depends greatly of the magnitude of the hyperbolic excess velocity. Below are a couple of tables with the Δv data. The first table is for V_∞ = 900 m/s, about that required for a trip to Duna, and the second table is for V_∞ = 2800 m/s, about that required for a trip to Jool. I think the columns titles are self explanatory but, just to be sure, here's an explanation:

Column 1 is the initial orbit altitude, assumed circular.

Column 2 is the Δv required to drop the periapsis altitude to 75 km; the apoapsis remains at the initial orbit altitude.

Column 3 is the Δv required to eject the spacecraft as it passes through periapsis of the eccentric orbit.

Column 4 is the total of columns 2 and 3, i.e. the total Δv required to perform the two burn ejection.

Column 5 is the Δv required to perform the ejection in a single burn from the initial orbit; compares to the total in Column 4.

In the first case (low V_∞) we see that the break even point is somewhere a little over 8,000 km. In the second case (high V_∞) we see that the break even point is right about 300 km. Interestingly, these altitudes are almost exactly the same transition points that I found in my opening post, where the Δv transitioned from decreasing with increasing altitude to increasing with increasing altitude (i.e. the low points on the graphs). In the OP I estimated that the transitions points occurred at 8,145 km and 303 km respectively, which concurs nicely with the data below.

It is starting to look like this transition point might have some profound implications. I must study this some more.
 

Initial Altitude (km)	(A) Lower Pe to 75 km (m/s)	(B) Ejection burn from 75 km (m/s)	Total (A) + (B)	Ejection burn from high orbit (m/s)
11,400	365.4	210.2	575.6	640.1
10,000	377.5	221.2	598.7	637.8
9,000	386.7	230.9	617.6	636.8
8,000	396.3	242.8	639.1	636.4
7,000	406.3	257.6	663.9	637.2
6,000	416.4	276.6	693.0	639.7
5,000	425.8	301.8	727.6	645.1
4,000	432.9	336.9	769.9	655.3
3,000	433.9	389.2	823.1	674.5
2,000	417.2	475.4	892.6	712.5
1,000	341.2	644.9	986.1	800.0
500	229.2	811.2	1040.3	897.3
300	146.9	912.4	1059.3	961.5
100	20.5	1049.6	1070.1	1055.4

 

Initial Altitude (km)	(A) Lower Pe to 75 km (m/s)	(B) Ejection burn from 75 km (m/s)	Total (A) + (B)	Ejection burn from high orbit (m/s)
11,400	365.4	1130.8	1496.3	2360.7
10,000	377.5	1141.8	1519.3	2339.4
9,000	386.7	1151.6	1538.2	2321.9
8,000	396.3	1163.4	1559.8	2302.2
7,000	406.3	1178.2	1584.6	2279.6
6,000	416.4	1197.2	1613.6	2253.5
5,000	425.8	1222.4	1648.2	2222.7
4,000	432.9	1257.6	1690.5	2185.7
3,000	433.9	1309.9	1743.7	2140.4
2,000	417.2	1396.1	1813.3	2083.6
1,000	341.2	1565.5	1906.7	2015.0
500	229.2	1731.8	1961.0	1984.6
300	146.9	1833.0	1980.0	1979.9
100	20.5	1970.3	1990.8	1988.3

 

Edited April 9, 2016 by OhioBob

[mhoram]

Here are my results about when to use a single-burn vs two-burns. (More details on original post)

[GoSlash27]

OhioBob,

I haven't thanked you for this work yet, so thanks! I think you've just come up with the most organized way to look at the problem.

Best,

-Slashy

[Red Iron Crown]

My experience with this:

- If launching from KSC it is better to use as low a parking orbit as practical, any savings from burning at higher altitude are negated by the expenditure to get there.

- If starting from orbit around Minmus it is almost always better to drop to low Kerbin Pe before ejecting.

- If using Minmus-harvested fuel for refueling it is better to bring the fuel to the ship in LKO rather than bring the ship to Minmus.

[Starhawk]

It is starting to look like this transition point might have some profound implications. I must study this some more.

Profound, indeed! This is fascinating stuff.

Not simple black and white at all.

Thanks to everybody contributing to our understanding.

Happy landings!

[Norcalplanner]

OhioBob,

I haven't thanked you for this work yet, so thanks! I think you've just come up with the most organized way to look at the problem.

Best,

-Slashy

OhioBob, I agree with Slashy 100%. These tables rock.

[NecroBones]

These tables are great! Nice.

Generally speaking, I think if we're talking about an escape burn, doing so at a lower altitude still makes a lot of sense. Even if 300 km saves a little (11 m/s or so) versus a 70 km parking orbit, it's going to be cheaper on the ascent toward your parking orbit to aim for a lower altitude anyway, so it may be a bit of a wash.

In any case, this is interesting to think about. I've had "course corrections" several times in how I think about orbital trajectories and what is more efficient, often because of someone much smarter or more knowledgeable than me pointing out things I hadn't considered. So I love seeing this stuff.

[Red Iron Crown]

I've had "course corrections" several times in how I think about orbital trajectories and what is more efficient, often because of someone much smarter or more knowledgeable than me pointing out things I hadn't considered.

So much this. E.g. I used to do plane changes by keeping the ship pointed at normal/anti-normal and rotating through the burn, until someone demonstrated to me that it is less efficient than a straight burn.

There's always more to learn.

[GoSlash27]

Ohio Bob,

I took the liberty of reverse- engineering your process here in a spreadsheet and everything seems to line up... except that my intercepts are still coming in under- budget. I can't figure out why

It wasn't all for naught though; I was able to recalculate the V∞ to reflect the actual excess velocity on a Hohmann transfer to each body*. This gave results that were still a little behind what I'm seeing in- game, but lines up precisely with all bodies in the subway map in the wiki.

Body/ V∞/ÃŽâ€V at 70km transfer/Calculated optimal altitude/ÃŽâ€V at optimal altitude

Moho/2,349/1,712/680km/1,661

Eve/779/1,043/11.0Mm/551

Duna/918/1,078/7.70Mm/649

Dres/2,088/1,564/1.06Mm/1,477

Jool/2,713/1,935/360km/1,918

Eeloo/2,955/2,094/220km/2,089

It looks like a transfer to Eve would show the discrepancy best. I'll try it tonight.

Best,

-Slashy

*Method: Excess velocity is simply what would be the DV on the first burn of a Hohmann transfer from one body's orbit to another. I calculated that using both Kerbin's and the target's semi- major axes.

Edited October 7, 2015 by GoSlash27

[OhioBob]

Alright, I think I have this figured out. As we know, when a spacecraft is ejected on a hyperbolic trajectory out of a planet's space, it is accelerate to beyond escape velocity. We can breakdown the ejection Δv into two components: (1) the part that gets us from initial orbital velocity to exactly escape velocity, plus (2) the part that is in excess of escape velocity. I will denote these two components Δv_esc and Δv_∞, respectively. We have,

Δv = Δv_esc + Δv_∞

Δv_esc is simply the difference between escape velocity and orbital velocity. The relationship between escape and orbital velocity is such that Δv_esc is always equal to 41.42% of orbital velocity. Since orbital velocity always decreases with increasing altitude, so does Δv_esc.

The second component, Δv_∞, is the part that is subject to the Oberth effect. Δv_∞ always increases with increasing altitude. It is the part that we minimize by burning deep in the planet's gravity well.

Our total ejection Δv, therefore, is made up of two components that trend in opposite directions. When we add the two components together, we get curves that look like those in the graphs in my opening post. There is a point at which the total Δv reaches a minimum value. If Δv_∞ is small in comparison to Δv_esc, then the minimum point will be far from the planet. If Δv_∞ is large in comparison to Δv_esc, then the minimum point will be close to the planet.

As noted in the OP,

Δv = (v_∞² + v_esc²)^0.5 - v_orb

which can also be written,

Δv = (v_∞² + 2μ/r)^0.5 - (μ/r)^0.5

where v_∞ is the hyperbolic excess velocity, μ is the planet's gravitational parameter, and r is the radius of the orbit.

Since the minimum point on the Δv vs. altitude curve is the point at which the slope is equal to zero, we can take the first derivative of the Δv equation, set it equal to zero, and solve for r. I'll spare you the math, but the end result is

r = 2μ/v_∞²

If we want the altitude, z, rather than the orbital radius, we simply subtract the planet's radius, r_o.

z = 2μ/v_∞² - r_o

For Kerbin, μ = 3.5316x10¹² m³/s², and r_o = 600,000 m.

The above equation gives the altitude at which the ejection Δv will be at its minimum value for a given v_∞. Of course, as has been noted numerous times already, in most cases what is most important is the total Δv needed to get from the surface of Kerbin to ejection. In this case the additional Δv required to reach higher orbits negates any potential savings. Parking orbits should be as low as practicable.

However, there are some situations where the above information might have a practical use. For instance, we've already seen that the altitude computed above defines a transition point, above which a two-burn ejection is most efficient, and below which a one-burn ejection is most efficient. (See post #31 for further explanation.) If there are any other practical uses for the data, then that's what you guys can figure out. I'm just here to provide the underlying math.

Of course to compute the altitude of minimum Δv, one must know v_∞. Hyperbolic excess velocity is not something that is commonly known, though it can be computed from readily available data. The method I typically use is to jot down data from Alex Moon's Launch Window Planner and plug it into the following equation,

v_∞ = [ ((μ/r)^0.5 + Δv)² - 2μ/r ]^0.5

For example, let's say that, for a trip to Jool, the launch window planner says the ejection Δv is 2000 m/s from a 100 km orbit. We have,

v_∞ = [ ((3.5316x10¹²/700000)^0.5 + 2000)² - 2*3.5316x10¹²/700000 ]^0.5 = 2818 m/s

We can then compute the optimum ejection altitude,

z = 2*3.5316x10¹² / 2818² - 600,000 = 289,446 m

Let's say I'm docked at a refueling station at an altitude of 500 km. Since I'm above 289 km, I should perform a two-burn ejection, dropping my periapsis to just above Kerbin's atmosphere and performing my final ejection at periapsis. (Of course I'd have to decide if the Δv savings are worth the trouble of the two-burn method.)

10/30/2015 - edited to add:

It has been discovered that the orbit whose radius is given by the formula r = 2μ/v_∞² is called a gate orbit.

 

Edited April 9, 2016 by OhioBob

[OhioBob]

mhoram said:

Here are my results about when to use a single-burn vs two-burns. (More details on original post)

Below is what I came up with using my equation,

z = 2μ/v_∞² - r_o

Of course my graph uses "hyperbolic excess velocity" rather than "velocity at SOI", though they are effectively the same thing. Above 100 km our curves look to be almost identical. Below 100 km I should have stopped my curve like you did - it makes no sense to continue it below the boundary of the atmosphere.

 

 

Edited April 9, 2016 by OhioBob

[GoSlash27]

OhioBob,

I just finished up my test, and this concept is definitely sound. *Proof of concept dance*

With the re-jiggered v_{∞}values I provided, here's the results:

Predicted DV to Eve intercept at 70km: 1,043 m/sec.

Actual DV to Eve intercept at 75km: 1,035 m/sec

Predicted DV to Eve intercept at 11Mm: 551 m/sec

Actual DV to Eve intercept at 11Mm: 537 m/sec

Still a hair high on the prediction, but pretty darn close.

Why do I keep coming in under the estimated value??

Best,

-Slashy

[OhioBob]

OhioBob,

I just finished up my test, and this concept is definitely sound. *Proof of concept dance*

With the re-jiggered v_{∞}values I provided, here's the results:

Predicted DV to Eve intercept at 70km: 1,043 m/sec.

Actual DV to Eve intercept at 75km: 1,035 m/sec

Predicted DV to Eve intercept at 11Mm: 551 m/sec

Actual DV to Eve intercept at 11Mm: 537 m/sec.

Still a hair high on the prediction, but pretty darn close.

Those are great test results. Thanks for taking the time to do it.

Why do I keep coming in under the estimated value??

I'm guessing it might be due to the game's patched conics. We don't have to actually reach escape velocity to escape, we just have to reach the SOI. That should mean that the prediction overestimates the amount of DV required.

[GoSlash27]

Ohio Bob,

Not a problem; it's science!

As long as I have a probe in the neighborhood, is there any other intercept you'd like empirical data for? I'm sitting at 11 Mm with 1,179 m/sec DV.

I've also got a probe sitting at 75km altitude.

Best,

-Slashy

[OhioBob]

Something else worth mentioning is that the mathematics governing orbit insertions is exactly the same as orbit ejections. Although all my examples talked about ejection Δv, the formulas and methods work just the same for orbit insertion. The only difference is the direction of travel. Therefore, if you are doing something like placing a station in orbit around another planet (where orbit altitude doesn't matter), the information presented in this thread can be use to select the most Δv friendly orbit.
 

Edited April 9, 2016 by OhioBob

[sardia]

My experience with this:

- If launching from KSC it is better to use as low a parking orbit as practical, any savings from burning at higher altitude are negated by the expenditure to get there.

- If starting from orbit around Minmus it is almost always better to drop to low Kerbin Pe before ejecting.

- If using Minmus-harvested fuel for refueling it is better to bring the fuel to the ship in LKO rather than bring the ship to Minmus.

It's better to have a refueling station in LKO instead of minmus orbit? The delta v savings outweigh the free delta v you got from burning to minmus and refueling there? It's so little delta v to burn into the solar system. Can you explain?

[Red Iron Crown]

It's better to have a refueling station in LKO instead of minmus orbit? The delta v savings outweigh the free delta v you got from burning to minmus and refueling there? It's so little delta v to burn into the solar system. Can you explain?

There is no station, only tankers. My strategy for this is:

1. Send full tanker from Minmus to LKO.

2. Rendezvous and dock with interplanetary ship.

3. Raise Ap to about Minmus altitude.

4. Refill interplanetary ship and undock tanker.

5. Complete interplanetary ejection (very cheap since Ap is already out at Minmus and there's a low Pe for maximum Oberth effect).

6. Fiddle with tanker orbit to intercept Minmus.

7. Capture at Minmus, reload, repeat.

It's pretty much equivalent to refueling the ship after ejecting from Minmus and setting a low Kerbin Pe for the escape burn, but much easier to time as I don't need to worry about Minmus' position as the window approaches. Step 6 is sometimes a bit costly, but not overly bad since the tanker uses efficient propulsion as it is intended to be reused so much (and that efficient propulsion can help with the step 3 burn of the IP ship).

I haven't fully processed the math that OhioBob and Slashy are talking about (and they are smart people to whom I pay close attention), so there might be room to improve this. I have been working under the assumption that the lower the Pe the more efficient the escape.

[cantab]

My experience with this:

- If launching from KSC it is better to use as low a parking orbit as practical, any savings from burning at higher altitude are negated by the expenditure to get there.

- If starting from orbit around Minmus it is almost always better to drop to low Kerbin Pe before ejecting.

- If using Minmus-harvested fuel for refueling it is better to bring the fuel to the ship in LKO rather than bring the ship to Minmus.

Depending on mission profile the third point may be wrong. The delta-V to get anywhere from Minmus (via a low Kerbin PE of course) is less than to get there from LKO. A pit stop at Minmus therefore allows you to use a ship with a lower maximum delta-V and so the ship can have a lower dry mass - less tankage, possibly fewer or weaker engines too.

The mission profile you just described is of course not a simple LKO refuel. I think it rather amounts to adding a booster to your interplanetary ship that does much of the ejection burn, and then contriving to reuse that booster. The reduction in maximum delta-V requirement is a little better than with a refuel at Minmus, though only by 200 m/s or so.

[Red Iron Crown]

There's some other advantages to the profile I mentioned, too (though they aren't dV related as much). Since the tanker will rendezvous in LKO, I can use the IP ship's entire tankage for ascent as I don't need to keep any fuel for a Minmus transfer and capture, reducing the size/mass/cost/part count of the initial lift. There is also the timing issue, with this strategy it's like any ejection from LKO so I can hit a window +/-15 mins; with an ejection from Minmus there's only about 10 days out of 50 when Minmus is in a good enough spot to eject (and I need to account for the time to drop to low altitude, but I'm getting better at that).

I just figured that since the tanker was going back to Minmus anyway it made sense to leave the final refueling until enough of the burn for transfer was already done, then continue with the ejection.

[mhoram]

Below is what I came up with using my equation,

z = 2μ/v_{∞}² - r_o

Of course my graph uses "hyperbolic excess velocity" rather than "velocity at SOI", though they are effectively the same thing. Above 100 km our curves look to be almost identical. Below 100 km I should have stopped my curve like you did - it makes no sense to continue it below the boundary of the atmosphere.

http://www.braeunig.us/pics/KSP/1burn_vs_2burn.png

It is good to know, that you came to similar results.

I did not stop the graph below 100km ... It is just that below that altitude a two-burn strategy (with an 80km altitude Periapsis) makes ÃŽâ€V-wise almost no sense, so the graph points up to indicate a preference for single burn.