Yasmy

[I]Edit: Oops: I just saw the second page of responses. Missed it before writing this. Nice tests OhioBob![/I] <Removed a bunch of junk> [COLOR="silver"][SIZE=1]- - - Updated - - -[/SIZE][/COLOR] OhioBob, if you are up for some more tests, how about splitting the burn [url=http://forum.kerbalspaceprogram.com/threads/125413-Precision-Burns-Before-the-Node-After-the-Node?p=2020522&viewfull=1#post2020522]like this instead[/url]: t = T (m/mf) (1 - sqrt(1 - mf/m)), where T is the total burn time, mf is the fuel consumed, and m in the initial total mass. Use the theoretical values for an instantaneous impulse burn. This should roughly split the delta-v in half before and after the node. I don't think it is the best split, but I suspect it is better than T/2. I expect that it will reduce the error in the longitude of periapsis. On the other hand, it would increase the periapsis. I'm interested in if it puts you closer to your intended orbit. Edit: Here is an equivalent, but perhaps more usable form: t = T (1-exp(-dv/(2 Isp g0))) / (1-exp(-dv/(Isp g0)). Then from your numbers, t = 0.571 T = 178.7 seconds, instead of T/2 = 156.2 seconds.

[quote name='Kraken1950']<snip> What do you mean by μ Kerbol, what value is it ? Mass x Gravitational constant ? Secondly, in the Hohmann transfer orbit formula why don't we use R (radius of parent body) in this formula ? Once out os Kerbin SOI the R could be Kerbol ? I will check on the vis-viva formula as do not understand the use right now. Yes, by mu (the English spelling of the Greek letter μ), I mean the gravitational parameter, G*M. It's easier to type than looking up μ and cutting and pasting, if you don't know a keyboard shortcut. M should be the mass of the object you are currently orbiting. In the Hohmann transfer orbit formulae, r1 and r2 are the distances from the center of the body you are orbiting at departure and arrival because these distances determine your gravitational potential energy. For interplanetary transfers, the central body is Kerbol, and r1 and r2 are approximately the orbital radii of the initial and final planets. Really they should be the spacecraft's Kerbol radii at the SOI boundaries, but the planetary orbital radii are a fabulous first approximation. The error is less than Kerbin's SOI radius divided by Kerbin's orbital radius: 8.42e7/13.6e12 = 6e-6. Completely ignorable. (For Jool the error can be on the order of 3%, since its SOI is so huge.) The way I used the vis-viva equation is entirely equivalent to using conservation of energy: E1 = 1/2 m v1^2 - GMm/r1 E2 = 1/2 m v2^2 - GMm/r2 E1 = E2, thus v2^2 = μ(2/r2-2/r1) + v1^2 Don't forget that in KSP, orbital altitudes are reported, which are distances from nominal planetary surface, rather than distances from the center of the planet, so you often have to add the planet's radius to its altitude to find r, the distance of the spacecraft from the center of a planet. So for transfer from LKO at 100 km altitude to geosynchronous orbit at 1581.76 km altitude, you would use r1 = 100 km + 600 km, and r2 = 1581.76 km + 600 km, since Kerbin's radius is 600 km. In contrast, the wiki lists planetary orbital information using semi-major axis, periapsis and apoapsis all measured relative to the center of Kerbol, not its surface.

I'll address your second question on how to calculate the cost of interplanetary travel. First, there are some high quality delta-v maps around. Be careful to use an up to date map. While I've calculated everything myself from time to time, I still use delta-v maps when planning most of my missions. Second: It appears you correctly calculated the first burn of a Hohmann transfer from low Kerbin orbit to the Mun, which is about 860 m/s. Next you might want to calculate the second burn to circularize in Munar orbit, but the second Hohmann burn equation is not correct because it ignores the presence of the Mun. The equations are [img]https://upload.wikimedia.org/math/1/b/5/1b54b322a2dd6656ea92162ac419bb28.png[/img] and [img]https://upload.wikimedia.org/math/d/7/f/d7f826b60051adf001ccd0f45159a618.png[/img] respectively. Now you can use these equations for part of an interplanetary travel calculation, but it is a bit more complicated than a Hohmann transfer orbit between two orbits around a single body. First though, you need to use mu for Kerbol, not Kerbin, and r1 and r2 as the orbital radii of your starting and ending planets at time of departure and intercept. So step 1 is calculating the Hohmann transfer burns. These are not the burns you will be performing. These are the burns you would make to go from one planet's orbit to another planet's orbit if the planets were not there. Next you have to account for the gravity of the planets. This is good, as the planets will actually help make the transfer cheaper. delta-v1 is how fast you want to be going right before exiting Kerbin's SOI, so that your Kerbol-relative velocity will be Kerbin's orbital velocity plus delta-v1. Likewise delta-v2 is how fast you will be moving relative to your destination planet when you reach it. Step 2 is to calculate how to exit Kerbin's SOI with velocity delta-v1, starting from some initial orbit around Kerbin. The easiest way is to use the [url=en.wikipedia.org/wiki/Vis-viva_equation][i]vis-viva[/i][/url] equation: [img]https://upload.wikimedia.org/math/4/0/6/4066c272245ac8455dde64dedf1d7bed.png[/img]. Now that we are back in Kerbin's SOI, use Kerbin's mu. While your velocity and radius change along an orbit, the semi-major axis is constant. We can use the vis-viva equation at two locations, exploiting this fact. At the SOI, the radius is the SOI radius, R, and the velocity is delta-v1. Now pick some initial low Kerbin orbital radius (call it r0) at which to perform your initial transfer burn. 1) (delta-v1)^2 = mu (2/R-1/a). Solve for the unknown semi-major axis: mu/a = 2mu/R - (delta-v1)^2. 2) Substitute the semi-major axis back into the equation at LKO: v1^2 = mu(2/r0 - 1/a) = mu(2/r0-2/R) + (delta-v1)^2 3) v1 is how fast you want to be going at LKO at radius r0, to ensure that you will be going delta-v1 at the SOI. Thus your first burn is delta-v0 = v1 - v0, where v0 is your initial LKO velocity. If your initial orbit is circular with radius r0, the initial orbital velocity is v0 = sqrt(mu/r0). (You can show this from the vis-viva equation.) Now you know your transfer burn: delta-v0 = v1 - v0. Be careful to use the correct mu everywhere. Kerbol's in delta-v1 and delta-v2. Kerbin's in v1. Step 3 is just like step 2. Given the Hohmann transfer arrival velocity delta-v2 at your destination planet's SOI, calculate the delta-v, delta-v3 needed to slow down to orbital velocity at some radius r3. The math is identical. Use the destination planet's mu in the vis-viva equation. Step 4. The total transfer and injection delta-v is delta-v0 + delta-v3, though if your destination has an atmosphere, you can reduce delta-v3 by aerobraking. (Aside: Some people burn so they just escape a planet, burn the first Hohmann transfer maneuver, (skip slowing down at the destination SOI) then burn to circularize in low orbit. For these maneuvers, the cost is (v_escape-v0) + delta-v1 + delta-v3, which is always higher than delta-v0 + delta-v3. You can prove this with the equations above. v_escape is the velocity needed at LKO to have zero velocity at the SOI, instead of delta-v1.) P.S: Calculating ejection angles is a good bit trickier. It is just geometry, but it is a pain to get right. Writing that up would take me some time. P.P.S: Calculating landing and liftoff delta-v is also way too much to cover in this single post. P.P.P.S: LKO to the Mun: use delta-v1, the LKO to Munar orbit transfer burn plus delta-v3, the circularization inside a destination SOI burn to calculate the cost of transfer from LKO to the Mun. As always, watch your mus.

Don't mouse-click anywhere? Doesn't always work. You might like to use a maneuver node editor instead, such as [url=http://forum.kerbalspaceprogram.com/threads/47863-1-0-5-Precise-Node-1-2-1-Precisely-edit-your-maneuver-nodes]Preciese Node[/url]. This stays open and allows very fine control of maneuvers. Great mod.

[quote name='More Boosters']Your gravitational potential energy is calculated very simply as m*g*h, and radial burns exist to change your altitude.[/QUOTE] This is only approximately true, useful only for very small height changes relative to the distance to the center of the body you are orbiting. Otherwise you should use the full form: PE = - GMm/r. For very small changes in height, h = dr << r: PE = -GMm/(r+dr) ~= -GMm(1-dr/r)/r = -GMm/r + GM/r^2 m * dr = -GMm/r + m g h, where g is the local acceleration of gravity, g = GM/r^2. Then the change in PE as a function of height is mgh, using the local g. But for orbital mechanics, you usually can't use this approximation.

[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] Short answer: Yes. Long answer: It is your sun-relative energy that matters for orbits in or crossing the sun's SOI. And for any particular solar orbit, your total sun-relative energy is the same no matter how you got on that orbit. (Not strictly true in KSP due to Kerbin's SOI system, which is not a conservative system, but close enough.) Setup: in LKO burn so you push your orbit to just barely escape Kerbin's SOI. Case 1: Wait until you exit Kerbin's SOI, then burn dv. Case 2: Burn dv immediately. Arrange in either case to exit Kerbin's SOI along Kerbin's solar prograde. Note, since F = m*a, and F = -G*M*m/r^2, the acceleration due to gravity (a = -G*M/r^2) is independent of mass, i.e. objects of different mass fall at the same rate. Thus we can ignore mass. Instead of energy, I'll use "specific" energy, which is just e = E/m. Case 1: Initially near Kerbin, e = 1/2 v^2 - GM/r^2 = 0. Your Kerbin-relative energy is zero. This is what it means to have just enough energy to escape Kerbin. Exiting Kerbin's soi, relative to the sun, you have Kerbin's velocity V, and are at Kerbin's orbital radius R. Now add dv to your velocity, and add in the sun's potential: e = 1/2 (V + dv)^2 - G Ms / R^2, with Ms equal to the mass of the sun. Case 2: Initially near Kerbin, e = 1/2 v^2 - GM/r^2 = 0. Your Kerbin-relative energy is zero. Add dv to your velocity: e = 1/2 (v+dv)^2 - GM/r^2 = [1/2 v^2 - GM/r^2] + v dv + 1/2 dv^2. Note the bracketed term is zero. e = v dv + 1/2 dv^2. (Aside: This is the Oberth effect: de/dv = v) When you exit Kerbin's SOI, you retain that excess energy as kinetic energy: e = 1/2 Ve^2 = v dv + 1/2 dv^2. Ve = sqrt(2 v dv + dv^2), which gets added to Kerbin's solar velocity V: e = 1/2 (V + Ve)^2 - G Ms/R^2 Now Ve is always greater than dv, so right now you know that you got more bang (Ve) for your buck (dv) at LKO that in solar orbit (dv). Now to make it clear which case is better, take the difference of the total sun-relative energies, case 2 - case 1. de = (1/2 (V + Ve)^2 - G Ms/R^2) - (1/2 (V + dv)^2 - G Ms / R^2) de = 1/2 V^2 + V Ve + 1/2 Ve^2 - 1/2 V^2 - V dv - dv^2 de = V Ve + 1/2 Ve^2 - V dv - 1/2 dv^2 de = V Ve + 1/2 (2 v dv + dv^2) - V dv - 1/2 dv^2 de = V (Ve-dv) + v dv de = V dv (sqrt(1 + 2v/dv)-1) + v dv Sure it is a bit ugly, but it is strictly non-negative. It's never better to burn in solar orbit when coming from LKO, from a dv efficiency standpoint. In the limit of dv -> 0, de -> 0. In the limit of dv >> v, then we can make the good approximation that sqrt(1+2v/dv) ~= 1 + v/dv, giving de ~= v(V + dv). But for most transfers from LKO you can't simplify the expression. None the less we know it is positive and significant.

Take any orbit: elliptical, parabolic, hyperbolic. Doesn't matter. With engines off, orbital energy stays constant. As you move farther away from whatever you are orbiting, you slow down, losing kinetic energy. That kinetic energy loss is exactly balanced by gain in gravitational potential energy due to climbing out of the gravity well. So orbital energy stays constant. Thrusting, on the other hand, changes your velocity, but doesn't change the acceleration on your craft due to gravity. So thrusting does change your gravitational potential energy. Initial kinetic energy = 1/2 m v^2. Burn dv. Final kinetic energy = 1/2 m (v+dv)^2 (Ignoring for the moment the fuel mass and non-prograde burns, just to keep it simple.) So the change in energy is dE = 1/2 m (v+dv)^2 - 1/2 m v^2 = m v dv + 1/2 m dv^2 ~= m v dv. This is the Oberth effect: The change in orbital energy due to burning dv is proportional to v. Thus the faster you are moving, the bigger the change in energy. Therefore it is best to expend your fuel when you are moving faster, e.g., at periapsis. Like LN400 said, you can't directly compare velocity on solar orbit with velocity on Kerbin orbit, because Kerbin is moving relative to the sun. Your sun-relative velocity at Kerbin departure is larger than your sun-relative velocity when you escape Kerbin by approximately Kerbin's escape velocity. Edit: Ninjaed by Slashy's nice, succinct answer.

What you need to do is send along a little friend probe. Every rocket needs a friend probe. Nice, happy friend probe.

If you want a measure of the rocket's acceleration, use TWR. If you want a measure of the rocket's vertical acceleration, use the effective TWR. This depends on your current flight path, rather than just on your rocket mass and engines. The effective TWR (as defined by the wiki) is just the vertical portion of your acceleration in units of g. If your effective TWR is greater than 1, your vertical velocity will increase. If your effective TWR is less than 1, your vertical velocity is will decrease. (Technically, instead of g, you would need to use GM/r^2, but for most purposes, g is sufficiently close to GM/r^2 in atmosphere or low orbit.) - - - Updated - - - Nope. While you are correct that if your rocket engines are all angled by angle 'a' relative to their symmetry axis, your TWR will be reduced by cos(a), that's is not apparently the effective TWR as defined in the wiki. Personally this is what I would have called effective TWR, but the wiki disagrees with us.

1) Drop a 0 m/s maneuver node half way between Kerbin and Duna, or on the line of nodes if it is far enough away from Duna and you have a significant relative inclination. 2) Focus on Duna (in map mode, right click on Duna -> focus, or just hit Tab however many times it takes). Once you are focused, you can see what your encounter looks like. 3) Rotate the camera so you can see your encounter and the maneuver node. Mess with the maneuver node until you get the desired encounter. If tiny adjustments move your encounter by too much to manually fine-tune, move the maneuver node forward along the trajectory. 4) If your engines are too powerful to execute the maneuver with precision, temporarily limit their thrust. Or create a new maneuver closer to Duna. Edit: ninjaed by snark!

I have not tried this since before the offset tool, so maybe this won't work anymore, but try this: Place the part. Save the ship under one name. Offset the part in the direction you want. Save the ship under a second name. Compare the geometry specifications of the two parts in the two ship save files. Whatever coordinate changed, keep changing it and reloading it in KSP until it is where you want it.

Well, Kethane has the Kerbal Unreconstitutionator. If you need a tiny bit of fuel more than you need that Kerbal...

Yup, about 1100 m/s, plus some extra for a plane change, from low Laythe orbit back to Kerbin. If your current elliptical Laythe orbit is pointing in the wrong direction, you could just burn suboptimally in the correct non-prograde direction for departure and still have lots of fuel left when you get home.

I don't disagree with you. I can be overly pedantic myself, and it is always best to try to be clear here. I don't know if there is a standard term for this particular maneuver, though on these forums, some people call it an Oberth maneuver. I don't know if people call it that elsewhere, or if there is just no accepted term for it. I don't like that term because the Oberth effect is always in effect. All fuel-ejecting maneuvers are Oberth maneuvers. Some are just better than others. Anyway, I thought it was somewhat appropriate to call it a powered slingshot since you come in from a different SOI and end up on a hyperbolic orbit. I should be clear when I'm playing fast and loose with terminology.

Yes. Number 1 is what I'm calling a powered slingshot. Your terminology may differ. Heck, everyone's terminology may differ. I'm OK with that. Escape the moon, dropping your periapsis to the parent, and complete the burn at the parent. The delta-v cost is the combined cost of dropping down to the parent and executing the transfer burn. Yup. I created the table for two reasons: 1) To help understand the tradeoffs involved in different mining operation locations vs. fuel depots around the parent, and 2) To know when to transfer directly back to Kerbin from a moon, vs. dropping down to the parent to complete the burn.

Going from the Mun to Duna or Eve, you are better off just going directly from the Mun, and not sling-shotting around Kerbin. For other destinations, a Kerbin slingshot is more fuel efficient. Here's an approximate table comparing the cost of slingshots versus direct transfers, assuming no plane change maneuvers and circular orbits. While it is somewhat inaccurate in absolute terms, in comparison terms it should still be useful.

For future reference, look at the EVA section of Key_bindings in the wiki, or from the KSP main menu, select Settings, then Input to see and modify your key bindings.

I don't doubt that some people see some improvement for some activities with alcohol use, as stated earlier, because of how alcohol interferes with normal processes. I severely doubt it improves any motor function or thought processes in anyone. There are quite a few studies[citations needed] that show that no amount of alcohol in any of the test subjects improved either response time or muscle coordination/acuity. You may play pool better after a drink, but it's not because your mind or body functions better as a result of alcohol. As for the drunk driver, I suggest you never get in his car and step up and talk to him about the damage he could do to others and himself. And let's not forget that your ability to judge your performance is impaired by alcohol. That improvement you think you have under the influence may not actually be real. And finally, I'm not a doctor. So don't take my word. Speak to your doctor. You definitely don't want to be taking this kind of advice from the internets.

Open the debug menu: Alt-F12. Go to the contract section and select the contract. Hit the complete button. Alt-F12 again to get rid of the debug menu. The problem with the contract system is it can push you to work on one thing, while you want to do something else. If you want a contract gone because it is interfering with your enjoyment, get rid of it.

I'm afraid I don't have anything helpful to add, but back in 0.19.1, after a while I could never load my refinery or this would happen: I got a lot of use out of this fuel refinery, resupplying both Minmus and LKO ships. I was sad to see it die, as spectacular as it was.

Like I said, "I'm sure you know." Wasn't being critical, just pedantic for the possible benefit of anyone who hasn't had physics yet.

Forgive me for being pedantic. I'm sure you know this. Tonne is a mass unit, and is independent of the local gravity. Weight is the force on the mass due to the local gravitational acceleration. So tonnes are the same everywhere, but weight changes. 500 tonnes * 9.8 m/s^2 = 4.9 MN weight on Kerbin 500 tonnes * 16.8 m/2^2 = 8.35 MN weight on Eve Edit: fixed SI prefix. doh.

I'm keeping my Windows machine back at version 7, so I can't answer for Windows 8, but I have had similar problems in the past with programs which were installed with admin privileges but run without admin privileges and want to write save files to the install directory. Bits and pieces end up in C:/Users/<username>/AppData/Local or C:/Users/<username>/AppData/Roaming or something like that. Like MarvinKitFox said, use search to find them. So for my Windows KSP installs, I keep them all out of the Program Files directories, and instead have them all in C:\KSP\. E.g.: C:\KSP\0.14 C:\KSP\0.16 ... C:\KSP\1.0.2 C:\KSP\1.0.4 All the save data is safely stored in the install directory, because Windows doesn't consider this to be a directory owned by admin and used by multiple users who want their own set of saves, screenshots and whatnot.

Aiming is not that hard. For example, we can hit a retroreflector on the moon to measure the Earth-Moon distance. This is not to say that these experiments are easy. They are difficult, but definitely doable.

But it is this complicated. There are no analytic solutions in an atmosphere. You have to chug through integration/approximation methods for a set of coupled differential equations, either by hand or by machine. Computers (people with tables, adding machines, pencils and paper) were doing simulations like these long before governments had electronic computers.