Efficient Ways To Get To Dunna

[Nick kerman]

Hello I am a bit new to K.S.P and I am hopping some one could run me through the basics of getting to landing a probe on Dunna e.g  rocket design, using maths to calculate when to leave I do no theirs a bunch of websites that help you but I enjoy doing the homework of it all.

 If someone could tell me that would be AMAZING thanks heaps guys!  

Spoiler

 

 

[Warzouz]

You can try with Kerbal-X It's able to get there, land, return to orbit, but i's hard to get back to Kerbin. You'll need a bit more fuel.

For a one way trip, you can reuse your mun lander with little tweaks :  Add 2 drogues chutes and several regular chutes (depending on it mass). You may need to do a partially powered landing depending on your terminal velocity.

To get there

- Wait for Duna being before Kerbin with a phase angle of 43° (1/2 of a quarter)

- Launch to 80km orbit

- Set on the dark side of Kerbin, increase prograde to escape velocity. Try to have your escape velocity nearly parallel to Kerbon orbit. You can also tweak the location of the node so the apoapsis is maximum. Then tweak again you node to get a Duna encounter.

- Burn that

- Outside Kerbin SOI (preferably on AN or DN), you can do a small correction burn. Set Duna periaps to 70km. If you're luck to get an Ike encounter, try to pass in front of it, that'll slow you down freely.

- Anyway, burn retrograde around periapsis to capture and circularize. As an alternative, you can try aerocapture or combine everything.

 

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[Zhetaan]

As @Warzouz indicated, the greatest part of interplanetary transfer is getting the angle correct.

(Warning:  Wall of text)

To calculate this yourself, you need to start with orbital information for the planets involved.

Kerbin is easy; it's in a circular orbit of 13.600 Gm.

Duna is harder; it has an eccentric orbit that goes between a periapsis of 19.669 Gm and an apoapsis of 21.783 Gm.  If you draw a line connecting the two apsides, you get what is called the major axis and it is 41.452 Gm.  For orbital calculation, we're interested in the semi-major axis, which is simply half that value (or 20.726 Gm).  Note that all of this information is easily found or derived from easily found information in map mode.

Since Kerbin is in a circular orbit, its periapsis is its apoapsis is its semi-major axis.  As I said, Kerbin is easy.

Now, let's consider how we're getting to Duna.  Never mind that we don't have a good flight plan yet; let's assume that we're starting at Kerbin's orbit and going to Duna's orbit.  These orbits exist whether or not the planets occupy them (consider moving a satellite between a 100km to a 500 km Kerbin orbit, for example; you can do it even if there are no other sats in the sky), so the concept of moving between orbits doesn't require there to be anything at either the departure or arrival points:  we're only insterested in the path to get there.  The type of path we're going to use is called a Hohmann transfer orbit (apologies if you know this already) and it's based on the premise that the most fuel-efficient way to go from lower orbit A to higher orbit B is to set up an eccentric orbit such that its periapsis is on A and its apoapsis is on B (it isn't always the most efficient, but the exceptions are rare).  In other words, when you start, you're at A, and half an orbit later, you'll be at B.  Half an orbit after that, you're back at A.

How do we find the orbital parameters for the transfer orbit?  We use the departure and arrival orbits!  Our interplanetary ship is starting at Kerbin's orbit and ending at Duna's orbit, so the transfer orbit's periapsis and apoapsis will correspond closely to Kerbin's and Duna's semi-major axes.  Again, this is a rough estimate that assumes Kerbin and Duna's orbits are perfectly circular and perfectly aligned with one another, but they are fairly close to that ideal and so provide a good example.

Our transfer orbit's periapsis is Kerbin's orbit, i.e. 13.600 Gm.  Its apoapsis is Duna's orbit, i.e. 20.726 Gm.  The total major axis of this orbit is 34.326 Gm, so its semi-major axis is 17.163 Gm.

Now we need to know how long it will take to travel this orbit.  For this, we can use Kepler's Third Law, which states that the orbital period (time to orbit once) is related to two things:  the semi-major axis of the orbit and the gravity of the object it's orbiting (specifically, the equation is T_{orbitalperiod} = 2π * (A_semi-major³ / MG)^1/2.  The nice thing about Kepler's third law is that eccentricity doesn't matter:  all orbits of a given semi-major axis in a system will have the same period.  The nice thing about these orbits is that they all orbit the same thing (the sun) so the gravitational parameter (MG) is the same in all three cases, which means that you don't actually need to figure out the mass and gravity of the sun if you know the orbital period and semi-major axis of at least one of those orbits.

Kerbin's orbital period is 426 6-hour days.  For Kepler's laws, we need this in seconds (and Kerbin's year is not exactly 426 days long; but the remainder can be sucked up into the rounding error), so 426 (days/orbit) * 6 (hours/day) * 60 (minutes/hour) * 60 (seconds/minute) = 9201600 seconds/orbit.  Plugging in what we know and a bit of algebraic wizardry later (you mentioned wanting to work this stuff out for yourself, so I won't show the specifics), it works out that for any object orbiting the KSP sun, the orbital period equals the semi-major axis raised to the three-halves power (break out your scientific calculator) times a constant equal to 5.8017 x 10^-9.

To check the answer, we can use Duna's semi-major axis:  again, I'll leave the details to the student, but all told, it gives an orbital period of roughly 801.5 days, which agrees pretty well with Duna's actual orbital period of 801.6 days.

Now, let's try it on the transfer orbit:  with a semi-major axis of 17.163 Gm, the orbital period works out to about 13,045,000 seconds, or 603.9 6-hour days.  Just by looking at it, it makes sense: the transfer orbit should be somewhere between the other orbits.  However, this isn't the number we want:  the transfer takes place over half an orbit.  Therefore, we want half the orbital period, or 302 days.

Now we know how long we'll be in transit.  The next job is to find out when to leave so that when we arrive at Duna's orbit, Duna will be there.  For that, we need a little creative imagery:  imagine the solar system as a clock face, and let's put Kerbin at the six o'clock position at the time of departure.  We want to time our departure so that when our vessel arrives at the opposite point of its orbit (at the twelve o'clock position), Duna will be there.  The clock visual isn't the best because the orbits in KSP go anticlockwise, but so long as you understand which direction you're going, you'll be fine.

If we want to arrive at Duna when Duna is at the twelve o'clock position, and our trip there takes 302 days, then we want to start when Duna is 302 days away from the twelve o'clock position.  That should be obvious.  To get that, we need Duna's orbital period--but wait!  We figured that out as a check on our modified Kepler equation!  302 days out of 801.5 days (I'm using the calculated value; I'm assuming you don't want to use anything else for this) is .3768 of the orbit.  We're assuming Duna's orbit is circular (if it's very eccentric, it needs another layer of calculation), so .3768 of a circle is, out of 360 degrees, 135 degrees away from twelve o'clock.

There's no reference point for the twelve o'clock position on our solar system clock, however, so let's use what we have.  Kerbin is at the six o'clock position when we start, which is 180 degrees away from twelve o'clock.  Subtract off the 135 degrees of Duna's position with respect to twelve o'clock and we get 45 degrees for Duna's position with respect to six o'clock, i.e. Kerbin.  The fact that it is positive 45 degrees means that Duna needs to be 45 degrees ahead of Kerbin in its orbit when you launch--again, this should make sense.  Duna's orbit is higher, therefore slower, so for your faster transfer orbit to get to the same place at the same time, you need to give Duna a head start.

Set up a manoeuvre node when Duna is 45 degrees ahead of Kerbin and you should get the encounter.  It may take a bit of tweaking and will certainly require a mid-course correction, but it should work.

Needless to say, this is the quick-and-dirty method of getting encounters.  It works well for Duna because Duna's orbit is nearly circular, coplanar, and concentric to Kerbin's.  It works less well for other planets with different orbital parameters.  For example, even with Duna, the slight eccentricity and inclination make it so that some transfer windows are better than others because there are extra dimensions of alignment that you need to consider in order to get the absolute best efficiency.  For a better example, consider Dres.  Dres transfer windows recur about every 1.25 years; a window is available before the end of Year 1, but if you wait for the window in Year 3, you can save nearly 500 m/s of delta vee by taking advantage of a more favourable orbital alignment.  However, these extra-favourable alignments only occur about every decade, so you may want to just pack a little extra fuel instead of a lot of extra snacks.  Besides, overcoming the limitations of planetary orbital arrangement is part of why you play KSP in the first place--you learn nothing by sitting at home and time warping!