OhioBob

That's what I do too, except I still try to time my launch so that the plane change is reduced as much as possible. A mistimed launch and we could have a plane change of 30-, 45-, or even 60-degrees or more, depending on the inclination of the munar orbit. By properly timing the departure, it might be possible to get that plane change down to 10-degrees, maybe less. When confronted with a large plane change, I think it comes down to a choice of (1) elliptical capture, or (2) waiting for a better launch window. I've done it both ways, though I usually find it easier just to warp ahead until I see a geometry that's more to my liking. I then like to perform a single capture burn that includes the plane change.

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.

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.

When you arrive at Mun you want the plane of your incoming trajectory to be as close as possible to the plane of the munar orbit that you're trying to match. Fiddling with the maneuver node is part of the answer, but you also have to wait for the axis of the munar orbit to be properly aligned relative to Kerbin. If you arrive at the wrong time, you may not be able to get anywhere close to the same plane. However, waiting a day for Mun to advance further along in its orbit can dramatically change the situation. For instance, if the axis of the munar orbit is pointed in the direction of Kerbin then, 10 hours later after Mun has completed a quarter revolution, the axis will no be pointed to the left as seen from Kerbin. You just have to figure out when the timing is right so you arrive with the least amount of costly adjustments to make. You might also be able improve your approach by making a mid-course correction after your initial trans-munar injection. This is particularly true if you need a normal/anti-normal component to your trajectory. In that case, I usually like to make my initial injection just a prograde burn. After that I place a second maneuver node at the first AN/DN, perform whatever plane change is necessary, and refine the targeting.

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

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

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 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.

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, Vsoi2 = 2μ * (1/Rsoi - 1/Rbo) + Vbo2 to compute how much Δv we need to provide at 70 km to have 289.57 m/s left over after crossing the SOI. Vbo = (289.572 - 2*3.5316E+12*(1/84159286 - 1/670000))0.5 = 3247 m/s Δv = Vbo - Vorb = 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.

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

Yeah, I usually do all my computations using real life concepts, even if they don't exactly match KSP's way of doing things. In most cases the math is actually easier that way. And, as you say, it is often a little conservative, which is not a bad thing. I'm glad you've found it useful. I think my web site audience has gotten much bigger since the introduction of KSP.

We had a similar discussion to this some time ago in which I made the following post. Let's first consider real life n-body physics. Suppose we are launching from an altitude of 75 km, where Vesc = 3235 m/s, and we have V∞ = 900 m/s. Our required burnout velocity is, Vbo = (V∞2 + Vesc2)0.5 = (9002 + 32352)0.5 = 3358 m/s Now, with KSP's patched conics. We no longer think in terms of V∞, instead we just need to make sure that our velocity is 900 m/s relative to Kerbin when we reach the edge of the SOI. We set Vsoi = 900 m/s and solve for Vbo. Vbo = (Vsoi2 - 2μ * (1/Rsoi - 1/Rbo))0.5 = (9002 - 2*3.5316E+12*(1/84159287 - 1/675000))0.5 = 3345 m/s Therefore, in KSP we require 13 m/s fewer Δv than we would in real life. This number is comparable to what Slashy found in his experiments. I feel pretty confident that the difference between Slashy's numbers and the predicted numbers is due to KSP's patched conics.

All the math is described in post #40, however I'm not suggesting that you maneuver your spacecraft into that orbit just because the Δv is lower. The Δv you spend to get into the orbit can be more than what you save. If you are launching from the ground, you should seek the lowest practical orbit. However, there may be other applications (perhaps as yet undiscovered) where knowing the optimum altitude could be useful.

The equation to compute the ÃŽâ€v of a combined altitude and plane change is, where Vi is initial velocity, Vf is final velocity, and θ is the plane change angle. You can use this equation to compute the ÃŽâ€v with the plane change and without it. You can then compare the difference it to what the ÃŽâ€v would be if you performed the plane change at apoapsis of a highly eccentric orbit. This way you can decide which is cheaper ÃŽâ€v-wise. However, you will find that small plane changes, up to a few degrees, can be made almost for free when you combine them with the capture burn. Even if you plan to do the bulk of the plane change at high apoapsis, you should always try to do at least a few degrees at capture because it comes so cheap.* (ETA) * This technique is used in real life whenever a geostationary satellite is launched out of Cape Canaveral. Because of the latitude of the launch site, the satellite is initially launched into a low Earth orbit with an inclination of about 29 degrees. When they perform the burn to inject the satellite into the geostationary transfer orbit, they’ll take a few degree of inclination out of the orbit because they can do so for so little additional ÃŽâ€v, even though the velocity is very high at that point. Then, when they reach geostationary distance and the velocity is slow, they perform the bulk of the plane change in combination with the orbit circularization burn.

I just discovered something I had not previously noticed. In an earlier post I gave the equation, r = 2μ/v∞2 which is the radius at which the ejection/insertion Δv is minimum for a given value of v∞. The interesting part is if we rearrange this equation we get, v∞ = (2μ/r)0.5 Some of you may recognize that as the equation for escape velocity. Therefore, what we have found is that Δv is minimum when v∞ = vesc. The equation for v∞ is, v∞2 = vbo2 - vesc2 If we substitute vesc for v∞ and rearrange, we obtain vbo2 = 2vesc2 Recognizing that vesc2 = 2vorb2, we substitute and obtain vbo2 = 4vorb2 Taking the square root of each side, we have vbo = 2vorb or, expressed another way Δv = vorb We therefore see that ejection Δv is minimum when we are at such altitude that our ejection velocity is exactly double our initial orbital velocity. I am not entirely certain what the implications of that is, but I find it an interesting oddity.

It depends on the magnitude of V∞. The break even point at an altitude of 68,000 km is V∞ = 321 m/s. Therefore, if V∞ > 321 m/s, then drop your periapsis and perform two burns. If V∞ < 321 m/s, then just burn direct from your current orbit. The thing is, you can't get to any planet in the system for under 321 m/s V∞, so you are always going to be better off performing a two-burn ejection from 68,000 km. The only exception might be if you are traveling to another nearby asteroid, where V∞ might be < 321 m/s.

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.

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

Below is what I came up with using my equation, z = 2μ/v∞2 - ro 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.

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 Δvesc and Δv∞, respectively. We have, Δv = Δvesc + Δv∞ Δvesc is simply the difference between escape velocity and orbital velocity. The relationship between escape and orbital velocity is such that Δvesc is always equal to 41.42% of orbital velocity. Since orbital velocity always decreases with increasing altitude, so does Δvesc. 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 Δvesc, then the minimum point will be far from the planet. If Δv∞ is large in comparison to Δvesc, then the minimum point will be close to the planet. As noted in the OP, Δv = (v∞2 + vesc2)0.5 - vorb 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, ro. z = 2μ/v∞2 - ro For Kerbin, μ = 3.5316x1012 m3/s2, and ro = 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.

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

The numbers suggest that, in real life, the major planets (Eve, Kerbin, Duna and Jool) should be quite bright when observed from one another. Within the game, the brightness and visibility of planets is much less than what they should be. For example, from Kerbin a full Minmus should have an apparent magnitude of about -11, yet it can be difficult to even spot Minmus. I wrote a long winded thread about how the planets would look from Kerbin if it were a real life solar system. Here it is if anybody is interested: http://forum.kerbalspaceprogram.com/threads/130927-Kerbal-Astronomy-101 Of course the thread is just hypothetical and doesn't represent what can actually be observed within the game.

I frequently do the same thing. For instance, if I want an equatorial orbit around the target planet, I adjust my approach trajectory so that the periapsis lies in the planet's equatorial plane. I can, therefore, combine a plane change with the capture burn to end up in an equatorial orbit. Much cheaper ÃŽâ€v-wise than performing two burns.

While you were posting the above, I was writing a response that agreed with you (see above post #21).

If there is any practical application of it, I think it is likely to be realized if we are using a refueling station or something like that. If we are simply launching from the surface of Kerbin and ejecting from orbit, then I see no practical application. The extra Δv it takes to reach the higher orbit cancels out any benefit. If we just want a short term parking orbit, then I believe the lower the better.

If someone where to do those tests, I recommend they do it for Duna rather than Jool. As the graphs in the OP show, the effect is much larger and more obvious with Duna. With Jool, small imperfections in the way the maneuver node is set up could result in greater error than the effect we're trying to observe. It's also possible that the game's patched conic method could skew the results away from the purely hypothetical. Patched conics is not something I took into consideration at all when I did the math.