OhioBob

Because you wrote: "The only thing that will destroy a chute is drag force (assuming temperature isnt a problem)? That means the parachute can be deployed in upper atmosphere safely, but they will break off once air density increases and drag force increases?" which looked to me like you were asking if the increasing air density would increase the drag force and rip the parachutes off. If that wasn't a question, then why the question marks? You'll never be able to deploy a parachute at such an extreme altitude as 60 km. Parachutes have a minimum semi-deploy pressure, which is tweakable. The minimum pressure setting is 0.04 atmosphere for main chutes and 0.02 atmosphere for drogue chutes, which are also the default settings. On Kerbin these pressures occur at altitudes of 17,590 m and 21,204 m. On Duna these pressures occur much lower, at 4,448 m and 9,266 m respectively.

The drag force on the parachute peaks sharply when it first deploys. This is because the deployment of the chute results in a sudden change in the vehicle's drag area and drag coefficient, and a rapid lowering of vehicle's terminal velocity. There is a sudden increase in the drag force until the vehicle reestablishes terminal velocity. If the parachute is going to rip off, this is the time. If the chute survives the initial deployment, you should be good thereafter. As the vehicle descends under the parachute, it will steadily slow down as the air thickens, maintaining terminal velocity. The drag force on the parachute should remain approximately steady.

I agree. The safe deployment of parachutes should be based on drag force, not velocity. Drag force is proportional to air density times velocity squared. Duna air is about 1/8th the density of Kerbin air, therefore we should be able to deploy the parachutes on Duna at a velocity that is about 81/2 times the safe deploy velocity on Kerbin, which would be about 700 m/s. This needs to be fixed.

Apollo 13

I tried to do something like that a few months ago. What I did was to take each engine and add to it a nose cone and fuel tanks until it's Eve sea level TWR was 1.4. I then computed how much dV each configuration could produce. It's just a hypothetical experiment, but I figured it gave me some idea of the combined effect of ISP and TWR. Adding the Vector to my old results, I find that the best engines in order of performance are: (1) Mammoth, (2) Vector, (3) Aerospike, (4) Twin-Boar, and (5) Mainsail. The drop off in performance after these top 5 was so large that I wouldn't even consider any other engines.

[quote name='Archgeek']Though I'm curious just how much delta-v the Mun would impart to a near-stationary craft in its passing.[/QUOTE] Escape velocity. The same velocity it takes to escape is added when the craft falls in toward Mun.

The following are not the best of images, but I thought it was a pretty neat event to witness. I happened to catch Laythe eclipsing the Sun while my probe was descending through the atmosphere of Jool. Right next to Laythe is another moon, but I don't remember if it was Vall or Tylo. The descent almost lasted long enough for the second moon to also eclipse the sun, but my probe imploded just before first contact. The big dark spot is the heat shield of my probe.

You guys are responding to posts that are over two years old.

From my observations, I agree with this. I think that maintaining a fixed heading along the maneuver node tangent line is a reasonable representation of what happens in the game. The thing I don't fully understand, and therefore cannot simulate, is how the vector auto-adjusts. My simulation is comparable to aligning to the original maneuver node marker and then switching to attitude hold for the duration of the burn. This would surely result in the maneuver node marker drifting away from the ship's heading, most noticeably toward the end of the burn. That's correct. Since I don't know how the game computes the adjustment, there's no way I can simulate it. That is what Red Iron Crown suggested a couple days ago when he posted the following. My experiments simply show that he had it right.

If everything with the ejection is spot on other than the flight path angle, then we just need to make a slight change to the direction of the velocity vector. If we wait and make the correction immediately after leaving Kerbin's SOI, then, for the example given, I estimate that the delta-v would be about 49 m/s for the 'align to maneuver node' trajectory and about 134 m/s for the 'align to prograde' trajectory. In practice, however, it's not quite that simple. I usually just create a new maneuver node and figure out what burn I need to make to reestablish a good intercept with the target. As Yasmy said, there's no need to get on the originally planned trajectory, we just need to get on a trajectory that ends with the correct arrival conditions.

As to the why, perhaps the following image will help to illustrate it. This is just a graph that I made using Excel, so I couldn't get real fancy with labels, etc. Imagine that the planet Kerbin is centered on the origin of the axes (0, 0) with a radius of 600 km. The colored curved lines represent the trajectories that the spacecraft will follow as it leaves low Kerbin orbit and heads out into space. The dashed green line represents the theoretical orbit that the spacecraft would follow if the dV were applied instantly at the maneuver node. The maneuver node is located at the coordinates 0, -680. The red line represents the trajectory we follow by burning prograde. The blue line represents the trajectory we follow by burning to the maneuver node or, more correctly, to the right as the image is oriented. Both the red and blue trajectories start 30 degrees before reaching the maneuver node, with the burn time centered on the maneuver node. Not only do we want to exit Kerbin space with the correct velocity, but also in the correct direction. The correct direction is along the path of the dashed green line. You can see that when we burn prograde, we immediately start to push the vehicle farther and farther away from Kerbin, ending in a significant deviation from where we want to go. When burning in the direction of the maneuver node, we stay much truer to the desired trajectory. In fact, it looks to me like the location of the maneuver node on the Navball auto-corrects as the burn progresses, which likely keeps us even closer on course than this image represents. Of course, either trajectory can be made to approximately match the green line by adjusting the start time of the burn. In the case of the blue line, starting the burn a little sooner would move the outgoing path a little closer to the green line (this is what Yasmy suggests). In the case of the red line, we would want to start the burn later, though I'm not sure there's a simple formula to compute the start time. Since the dV loss that comes with burning to the maneuver node is relatively small, there seems to be little advantage to using the prograde method. Burning to the maneuver node appears to be the easier and better solution.

Sorry, guys. I just discovered a typo in my formula that computed the longitude. This changes the numbers. In the comparison between "align to prograde" versus "align to maneuver node," the latter still wins out as being the closest to the theoretical orbit, however the Δv has gone up slightly. In regard to the hybrid method, it is now appears to be the worst option in terms of accuracy. Please forget everything I said about it. Below are the revised numbers. Scenario #1, aligned to Prograde Scenario #2, aligned to Maneuver Node Scenario #3, aligned to Node then Prograde Theoretical Ideal Orbit Semi-major axis, km -450459 -450459 -450459 -450459 Eccentricity 2.58601 2.50234 2.54686 2.50957 Periapsis altitude, km 114432 76742 96799 80000 Longitude of periapsis, deg. -2.006 0.924 5.083 0.000 Longitude at infinity, deg. 110.74 114.48 118.20 113.48 Δv, m/s 2054.4 2076.4 2056.9 1990.4 In regard to Yasmy's method, using his equation to estimate the burn start time and aligning to the maneuver node now comes out very close in terms of accuracy. The method appears to warrant further study. Yasmy method, aligned to Maneuver Node Semi-major axis, km -450459 Eccentricity 2.50326 Periapsis altitude, km 77158 Longitude of periapsis, deg. -0.228 Longitude at infinity, deg. 113.32 Δv, m/s 2077.1 I'd like to stress again that when I say "aligned to maneuver node," what I'm really doing is maintaining a fixed attitude throughout the burn in the direction of a tangent drawn through the maneuver node. This may not be exactly the same as keeping aligned to the maneuver node on the Navball. When performing maneuvers in the game, the location of the maneuver node on the Navball seems to float around a bit, particularly toward the end of a burn. Since I don't know how the game computes the direction of the maneuver node marker, I can't exactly replicate it in my simulation. Therefore, in game results might be different than the results of my simulations. I'd like to also note that, in reporting my results, I've arbitrarily set zero longitude to be in the direction of the maneuver node. This is probably apparent but I should have mentioned it earlier. Sorry again about the error.

I think that any of the scenarios can be made to work well. It's just a matter of timing the burn so that the resulting departure is at the correct angle. However, I don't think we want a method that involves a bunch of complex math to determine when to start the burn. The common practice is to split the burn time 50/50 around the maneuver node, which is easy to do. In that case, following the maneuver node appears to get us closer to the desired trajectory and, therefore, should require a smaller course correction (though I have not done any calculations to determine the magnitude of the correction). This concurs with my in game experience as well, as little as there is. The idea of following the maneuver node for half the burn and then switching to the prograde marker I think requires more testing. It appears to work very nicely for this particular example, but I need to make sure that is not a fluke. I'd like to test the idea under different sets of circumstances to see if there is any consistency to the result.

*** NOTE THAT THESE RESULTS CONTAIN A MATHEMATICAL ERROR. PLEASE SEE REPLY #30 FOR CORRECTION. *** Using your equation I compute t = 178.5 s. This means is that I have to start the burn 34.3o prior to the maneuver node rather than 30o. Making this change modifies the numbers to the following: Scenario #1, aligned to Prograde Scenario #2, aligned to Maneuver Node Theoretical Ideal Orbit Semi-major axis, km -450472 -450467 -450459 Eccentricity 2.58596 2.51420 2.50957 Periapsis altitude, km 114430 82098 80000 Longitude of periapsis, deg. -11.806 -3.616 0.000 Longitude at infinity, deg. 100.94 109.82 113.48 Δv, m/s 2054.4 2071.8 1990.4 - - - Updated - - - I just discovered something interesting. On a whim I decided to see what happened if I aligned the thrust vector with the maneuver node for the first half of the burn, and then with the prograde marker for the second half of the burn. The resulting longitude of periapsis was a very close match to the theoretical orbit. I think this is the best result of the options tested so far. Perhaps we've found a better way to execute our burns? I think some in game testing is warranted. Scenario #3, aligned to Node then Prograde Theoretical Ideal Orbit Semi-major axis, km -450454 -450459 Eccentricity 2.54487 2.50326 Periapsis altitude, km 95892 77158 Longitude of periapsis, deg. 0.170 -0.228 Longitude at infinity, deg. 113.31 113.32 Δv, m/s 2058.1 2077.1

*** NOTE THAT THESE RESULTS CONTAIN A MATHEMATICAL ERROR. PLEASE SEE REPLY #30 FOR CORRECTION. *** I just tested the two scenarios using a simulation. The test was for a hypothetical trip to Jool. I figured an 80 km parking orbit and a theoretical instantaneous Δv of 1990.4 m/s. This Δv gives a hyperbolic excess velocity of 2800 m/s, which is typical for a Jool mission. I equipped the vehicle with a Poodle engine and gave it an initial mass of 51721 kg. This mass was chosen because it yields a theoretical burn time of 312.47 s, which is 1/6th of an orbit. I started the burn 30o before the maneuver node, i.e. half before the node and half after. In scenario #1 the thrust vector was maintained in a prograde direction throughout the burn. In scenario #2 the thrust vector remained locked in a fixed attitude in relation to the stars. The fixed attitude was that of the orbit tangent through the maneuver node. Scenario #2 should approximate keeping the vehicle aligned to the maneuver node marker on the Navball. In both cases I terminated the burn the instant the hyperbolic exess velocity reached 2800 m/s. In both cases the actual Δv ended up greater than the theoretical, which is what we would expect. However, the difference between the two scenarios was small. Keeping the vehicle aligned to the prograde marker was slightly better at 2054.4 m/s, versus 2067.4 m/s for the fixed thrust vector alignment. In the second scenario, the vehicle altitude dipped as low as 75.26 km. Although both scenarios resulted in the same hyperbolic excess velocity, the resulting orbits were different. Below is a comparison of the orbital parameters of the two scenarios and the theoretical ideal orbit based on the Δv being applied instantly at the maneuver node. As you can see, scenario #2, i.e. alignment to the maneuver node, is a closer match to the theoretical orbit. Of course, it is possible to adjust the start time of either scenario to improve the longitudes; however, if we stay with the practice of centering the burn on the maneuver node, then scenario #2 is better. Scenario #1, aligned to Prograde Scenario #2, aligned to Maneuver Node Theoretical Ideal Orbit Semi-major axis, km -450472 -450473 -450459 Eccentricity 2.58596 2.51543 2.50957 Periapsis altitude, km 114430 82660 80000 Longitude of periapsis, deg. -7.506 -2.466 0.000 Longitude at infinity, deg. 105.24 110.96 113.48

I don't have much experience with this, but I did once perform an experiment comparing the two methods. I didn't bother to compare the dV because I was more interested in finding out which method got me closer to the desired trajectory. (I figured that if the less accurate method were cheaper in terms of dV, the difference would likely be negated by having to perform a larger course correction.) My conclusion was that following the maneuver node was better, so that's what I've done ever since.

[quote name='Red Iron Crown']AFAIK the wiki is already fixed, is there some functionality that is not working?[/QUOTE] [COLOR=#222222][FONT=Verdana] It’s been awhile since I last tried the functions that were giving me problems, but the Wiki is still displaying the message: [/FONT][/COLOR][B][FONT=Verdana][COLOR=#000000]“This wiki is currently experiencing [/COLOR][URL="http://wiki.kerbalspaceprogram.com/wiki/Kerbal_Space_Program_Wiki:Migration_problems"][COLOR=#0000ff]migration problems[/COLOR][/URL][COLOR=#000000]. This is known and will be fixed at some point.”[/COLOR][/FONT][/B][FONT=Verdana][B] [COLOR=#000000] [/COLOR][/B][COLOR=#000000]If it is fixed[/COLOR][COLOR=#000000], then the message should be taken down.[/COLOR][B][COLOR=#000000] [/COLOR][/B][COLOR=#000000] The functions that were giving me problems were (1) couldn’t upload images, and (2) math text wouldn’t work.[/COLOR][/FONT]

Will this change also fix the Wikipedia?

[quote name='Choctofliatrio']So can we just get an accurate summary? What's the best periapse for Eve aerobraking/aerocapture? Is it high? Low? Medium?[/QUOTE] [FONT=Calibri][SIZE=3][COLOR=#000000]I think that for most cases the answer is about 65±5 km.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] It varies depending on the entry velocity and the drag characteristics of the vehicle.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] If your entry velocity is slow and/or you have a low ballistic coefficient, then set your periapsis on the high end. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]If your entry velocity is fast and/or you have a high ballistic coefficient, then set your periapsis on the low end.[/COLOR][/SIZE][/FONT][FONT=Calibri][SIZE=3][COLOR=#000000] It will vary case by case; you might have to experiment to find out what works best. [/COLOR][/SIZE][/FONT][FONT=Calibri][SIZE=3][COLOR=#000000] The equation for ballistic coefficient incorporates the drag coefficient, which is not easily determined.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] A substitute for ballistic coefficient is what I call the [/COLOR][/SIZE][I][SIZE=3][COLOR=#000000]shield loading[/COLOR][/SIZE][/I][SIZE=3][COLOR=#000000], which is the mass of the vehicle divided by the square area of the heat shield(s). [/COLOR][/SIZE][SIZE=3][COLOR=#000000]I ran experiments using test vehicles with shield loadings ranging from 600 kg/m[/COLOR][/SIZE][SUP][SIZE=2][COLOR=#000000]2[/COLOR][/SIZE][/SUP][SIZE=3][COLOR=#000000] to 8400 kg/m[/COLOR][/SIZE][SUP][SIZE=2][COLOR=#000000]2[/COLOR][/SIZE][/SUP][SIZE=3][COLOR=#000000].[/COLOR][/SIZE][SIZE=3][COLOR=#000000] And I tested entry velocities ranging from about 4600 m/s to 5000 m/s, which is typical for intercepts originating from Kerbin.[/COLOR][/SIZE][/FONT][SIZE=3][COLOR=#000000][FONT=Calibri] [/FONT][/COLOR][/SIZE][FONT=Calibri][SIZE=3][COLOR=#000000] The 65±5 km number worked well for most cases in the middle.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] Only at the extremes did I get outside that range. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]For example, when testing the highest shield loading in combination with the highest entry velocity, my ship would flip and explode. It was the only case in which I couldn’t complete a successful aerocapture.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] And when testing the lowest shielding loadings, a periapsis of about 70-75 km was needed.[/COLOR][/SIZE][/FONT]

[quote name='sdj64']Now, aerocapture at Eve is another story, one which almost always ends in a big fireball for any ship. Eve missions should plan for propulsive capture now.[/QUOTE] [FONT=Calibri][SIZE=3][COLOR=#000000]I did a bunch of aerocapture tests at both Eve and Jool and found little problem. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]It is entirely possible to aerocapture at Eve.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] The only serious problem I had was when I got down to an altitude approaching 60 km and the vehicle wanted to flip around. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]This was a stability problem and not a heating problem.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] In most cases I found that an intercept periapsis of about 65 km worked pretty well with neither heating nor stability problems.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] There should be little need to have a periapsis lower than that unless you have a very high entry velocity or a very high ballistic coefficient. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]If you have both a high entry velocity and a high ballistic coefficient, then you’ll likely burn up. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]If you have a vehicle with a very low ballistic coefficient, then you may need a periapsis closer to 70-75 km. [/COLOR][/SIZE][/FONT][FONT=Calibri][SIZE=3][COLOR=#000000]In any case, you’ll definitely need heat shields. [/COLOR][/SIZE][/FONT][FONT=Calibri][SIZE=3][COLOR=#000000] I also performed some Eve entry tests.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] Again my biggest problem was having the ship want to flip.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] Above 60 km there wasn’t much problem, but as soon as I got deeper in the atmosphere stability was a big issue.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] Activating RCS helped to stabilize it, but it still flipped when I got <50 km. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]I ended up adding aerobrakes at the top and that solved the problem.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] By the time I needed to deploy the aerobrakes I was slow enough that they didn’t overheat. [/COLOR][/SIZE][/FONT][FONT=Calibri][SIZE=3][COLOR=#000000]In fact, I didn’t have any problems with overheating as long as I could keep the ship stable. [/COLOR][/SIZE][/FONT][FONT=Calibri][SIZE=3][COLOR=#000000] By the way, I had the same flipping problem at Jool, which wanted to occur at about 150-155 km. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]Above that altitude, aerocapture was easy.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] For most cases a periapsis altitude of about 160 km worked well. [/COLOR][/SIZE][SIZE=3][COLOR=#000000]For very low ballistic coefficients, 165-170 km would be better.[/COLOR][/SIZE][SIZE=3][COLOR=#000000] Like Eve, a combination of high entry velocity and high ballistic coefficient should be avoided.[/COLOR][/SIZE][/FONT]

[quote name='Goody1981']Did you guys see my explanation?[/QUOTE] Yes. I thought it was an interesting explanation. A simple illustration of the concept.

[quote name='LostOblivion']Not sure if this was answered, or if his question was understood correctly, but is he right? For destinations in the inner solar system, does a sun-prograde burn out ahead of Kerbin, and then burn sun-retrograde, harness the Oberth effect better than a direct sun-retrograde burn?[/QUOTE] [FONT=Verdana][COLOR=#000000]As Slashy said, no.[/COLOR][COLOR=#000000] The best way to get to any planet, inferior or superior, is to perform a single burn close to Kerbin. [/COLOR][COLOR=#000000] If we accelerate to exactly escape velocity, we’ll have no velocity left over, relative to Kerbin, after escaping.[/COLOR][COLOR=#000000] We would effectively be in the same orbit as Kerbin, only leading or trailing the planet by a small amount. [/COLOR][COLOR=#000000]To reach another planet we would then have to perform a second burn to either speed up relative to the Sun, to reach a superior planet, or slow down relative to the Sun, to reach an inferior planet.[/COLOR][COLOR=#000000] Instead of performing the second burn, it is better to give ourselves a little extra velocity when we perform the escape burn so that we have some velocity left over after escaping.[/COLOR][COLOR=#000000] This left over velocity is called [/COLOR][I][COLOR=#000000]hyperbolic excess velocity[/COLOR][/I][COLOR=#000000], denoted V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000]. It is measured relative to Kerbin and it doesn’t matter whether it is prograde or retrograde. [/COLOR][COLOR=#000000]We want to give V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000] just the right value that, after we escape Kerbin, we are traveling at the correct velocity relative to the Sun that we will reach the target planet without having to perform a second burn. [/COLOR][COLOR=#000000] Let’s say we want to reach Eve.[/COLOR][COLOR=#000000] If we performed two burns, it takes about 950 m/s to first escape Kerbin. [/COLOR][COLOR=#000000]It then takes a second burn of about 750 m/s to modify our solar orbit to reach Eve.[/COLOR][COLOR=#000000] That’s a total of 950 + 750 = 1700 m/s. [/COLOR][COLOR=#000000] On the other hand, let’s say we perform a single ejection burn close to Kerbin. [/COLOR][COLOR=#000000]We have to give our spacecraft enough kinetic energy to first escape, plus enough additional kinetic energy that we have some leftover after escaping. [/COLOR][COLOR=#000000]We do this by accelerating to some [/COLOR][I][COLOR=#000000]burnout velocity[/COLOR][/I][COLOR=#000000], V[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][COLOR=#000000], whose kinetic energy is equal to the sum of the escape energy and the residual energy.[/COLOR][COLOR=#000000] That is, [/COLOR][COLOR=#000000] ½ mV[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] = ½ mV[/COLOR][SUB][COLOR=#000000]esc[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] + ½ mV[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000][SUP]2[/SUP] [/COLOR][COLOR=#000000] Canceling out ½ m, we have V[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] = V[/COLOR][SUB][COLOR=#000000]esc[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] + V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000][SUP]2[/SUP] [/COLOR][COLOR=#000000] Let’s say we’re in a 75 km parking orbit, where V[/COLOR][SUB][COLOR=#000000]esc[/COLOR][/SUB][COLOR=#000000] = 3235 m/s.[/COLOR][COLOR=#000000] We set V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000] = 750 m/s, which we determined earlier is the velocity we want to be traveling relative to Kerbin in order to reach Eve. [/COLOR][COLOR=#000000]Therefore, [/COLOR][COLOR=#000000] V[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][COLOR=#000000] = ( 3235[/COLOR][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] + 750[/COLOR][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] )[/COLOR][SUP][COLOR=#000000]1/2[/COLOR][/SUP][COLOR=#000000] = 3321 m/s [/COLOR][COLOR=#000000] Our initial orbital velocity is 2287 m/s, therefore the Δv is 3321 – 2287 = 1034 m/s. [/COLOR][COLOR=#000000]This is far less than the 1700 m/s it would take to perform the two burn strategy.[/COLOR][COLOR=#000000] Also note that the magnitude of the ejection burn is the same whether we burn prograde or retrograde. [/COLOR][COLOR=#000000]If we burn prograde, we’ll be going +750 m/s relative to Kerbin, and if we burn retrograde we’ll be traveling -750 m/s relative to Kerbin.[/COLOR][/FONT]

[COLOR=#222222][FONT=Verdana][quote name='KerikBalm']Curious... the MW of Eve's atmosphere is 43 according to the wiki, so it is more than 5x as dese as kerbins atmosphere at the surface now, right?[/QUOTE] Don’t forget that Eve is hotter too. Density is given by the formula, ρ = (PM)/(RT) where ρ is density (kg/m[SUP]3[/SUP]), P is pressure (Pa), M is molecular weight (kg/kmol), R is the universal gas constant (8314.4621 J/kmol-K), and T is temperature (K). Kerbin’s air pressure at sea level is 101325 Pa, the molecular weight of its air is 28.9644 kg/kmol, and its globally averaged temperature at sea level is about 287 K. Therefore, [/FONT][/COLOR] [COLOR=#222222][FONT=Verdana]ρ = (101325*28.9644)/(8314.4621*287) = 1.230 kg/m[SUP]3[/SUP] Eve’s air pressure at sea level is 506625 Pa, the molecular weight of its air is 43 kg/kmol, and its globally averaged temperature at sea level is about 408 K. Therefore,[/FONT][/COLOR] [COLOR=#222222][FONT=Verdana] ρ = (506625*43/(8314.4621*408) = 6.422 kg/m[SUP]3[/SUP][/FONT][/COLOR] [COLOR=#222222][FONT=Verdana] [quote name='KerikBalm']Many of the aerodynamic calculations actually account for atmospheric density, rather than simply pressure, as in older versions, correct?[/QUOTE] [/FONT][/COLOR][FONT=Verdana][COLOR=#000000]The aerodynamics forces of drag and lift are a function of density, not pressure.[/COLOR][COLOR=#000000] Through pressure does effect things such as the thrust of an engine.[/COLOR][/FONT]

[quote name='Tatonf']Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit. You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ? That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?[/QUOTE] Let's say you want to get to Duna. If you perform this transfer from solar orbit near Kerbin, it takes a Δv of about 920 m/s. But to get into that solar orbit you must escape Kerbin orbit, which takes about 950 m/s. That's a total of 1870 m/s. Suppose now that instead of performing two burns, you just give yourself enough extra velocity during your escape from Kerbin that you have some velocity left over after escaping. You can do it so that you are already traveling 920 m/s relative to Kerbin after escaping so you don't have to perform the second burn. The extra velocity that you need to provide when close to Kerbin is much smaller than the amount you retain after escaping. To retain 920 m/s requires an ejection burn of only 1080 m/s. This is far less than the 1870 m/s that it took to perform the two-burn strategy.

[quote name='KerikBalm']...[COLOR=#3e3e3e][FONT=Verdana] and Dunas is a mere 33% of what it used to be at the surface.[/FONT][/COLOR][/QUOTE] [FONT=Verdana][COLOR=#000000]This is true, but the molecular weight of the gas has tripled, so the air density is the same as before. [/COLOR][COLOR=#000000]For some reason the old atmosphere had a molecular weight of 14, which didn’t make any sense for such a small planet.[/COLOR][COLOR=#000000] It is now 42, but with 1/3 the air pressure. [/COLOR][COLOR=#000000]With the same air density, lift and drag should be unchanged. [/COLOR][COLOR=#000000] [/COLOR][/FONT][COLOR=#222222][FONT=Verdana][quote name='KerikBalm']By the way, do you know what game file specifies these things? I've heard there was a file that even specifies the average molecular weight of the atmosphere's gas... but I can't seem to find it[/QUOTE] There is a file that contains this data but, unfortunately, it is not readable as a separate config file.[/FONT][/COLOR]