Brainlord Mesomorph

RE-UPDATED: Long Burns: Interplanetary Trajectories with Low TRW (and No Math).

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Long Burns: Interplanetary Trajectories with Low TRW (and No Math)

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How to Perform the Mangalyaan Maneuver under Visual Flight Rules

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Finding Your Precise Interplanetary Ejection Angle with a Sundial

 

By Brainlord Mesomorph

 

“You don't need to do math for a long burn, just look at your watch, squint at the sun, and head off in the right direction" Jebediah Kerman :)
Note: I am not a physicist or a mathematician (but if you are, I'd like to hear from you).
EDIT: I HAVE heard from both, and they approve!

So you've been to this web page, http://alexmoon.github.io/ksp/ , and you've gotten (among other things) your departure window date and your Ejection Angle. The question at hand is: How do you use this information to plot your interplanetary course? And how do you get that big oversized, underpowered ship of yours on its way? This tutorial will attempt to answer both questions.

 

 

UPDATED: The technique has been refined with experience. Tutorial has been rewritten with data added for ALL planets in the Solar System. Optional Step 2B, provides for Exact Departure Time. Game screenshot showing how to plot the maneuver. AND THEN RE-UPDATED after the Forum upgrade ripped it a new one.

 


Step 1: Find the Ejection Angle

First, let's look at what the Ejection Angle means: The ejection angle is a point in your spaceship's orbit around Kerbin, where you should begin your interplanetary departure burn. Measured in degrees from prograde, or retrograde at the time of departure (T). For the purposes of this illustration our ejection angle is 122° from prograde.

 

 

KDO1.jpg

Now, by no coincidence, at that exact same point in Kerbin's orbit around its sun, the planet Kerbin, the sun, and the ejection point form a straight line.

 

KDO2.jpg

I call that the Solar Alignment Date. You can easily find the alignment date with the departure date and the ejection angle using the following formula:

 

D = (Ejection Angle -90) / X

Where D is the number of days prior to departure when the alignment takes place and X is how far the planet makes it around its Sun in one day.
Depending on the planet, and whether you use Earth time or Kerbin time, I've calculated all the values of X below.

X = Degrees around the Sun in one Day:
  Earth Time Kerbin Time
Kerbin 3.38 0.84
Moho 14.04 3.51
Eve 5.5 1.37
Duna 1.8 0.45
Dres 0.65 0.16
Jool 0.3 0.07
Eeloo 0.2 0.05
if D < 1, multiply by hours per day to get Hours before T
Hours/Day 24 6

 

So, given an ejection angle of 122° on Kerbin, your alignment date is T minus 9.6 days.

All you have to do is, on the Alignment Date, perform a maneuver to put your perigee and apogee points on a line with the sun.

KDOInGame.jpg

In our example we are going prograde, (away from the sun) so we will place our perigee point at the "midnight" point in our spaceship's orbit around Kerbin.

For retrograde trajectory as you would use the noon point. It's easy to remember with the following mnemonic:

 

"To go towards the sun, your perigee is towards the sun, to go away from the sun your perigee is away from the sun"

Its worth noting that the ONLY thing you have to on the Solar Alignment Date is
mark the Pe point WITH ANY SPACECRAFT you have. Not necessarily the one that's leaving. 

I actually keep a "NavSat" in LKO just for marking ejection angles of upcoming  flights that haven't been built yet.
 


Step 2: The Mangalyaan Maneuver

The first burn is the most important because it aligns your perigee and apogee points to the sun. After your done with that, your next step is the Mangalyaan Maneuver (sometime called "perigee kicking" in KSP forums), it's a series of prograde burns performed directly at the perigee point to raise the apogee. In the Kerbin system the highest we can raise the apogee conveniently is about 9000 km. Beyond that you run the risk of the Mun passing by and pulling your ship out of orbit.

I don't think it matters how many burns you perform or how long they are. If your ship can do it in one burn that' fine but you don't have too. I like to use a series of 6 minute burns because Kerbal alarm clock goes off 3 minutes before a node, and I burn until 3 minutes after the node. Then I wait until I come around again. (The total of all of the burns comes out to approximately 900 m/s) In the end, your orbit has a perigee in LKO, 75 to 80 km (not below 75), and apogee of around 9000 kilometers. (I call that Kerbin Departure Orbit, KDO) Take your time, in our example you have a full 9½ days to perform these maneuvers.

By the time your departure date (T) comes around your orbit no longer points at the sun and will look something like this. Your perigee point is your ejection point, and by no coincidence, also the point in your orbit when you have the maximum forward velocity, and when you're moving in exactly the right direction.

 

 

KDO3b.jpg

Now all you have to do plot your interplanetary burn directly at the perigee point , prograde (and about 900 m/s less than you were originally told.) You'll be surprised at how perfectly this course intersects your target plane's orbit. (I sure was!)

 


Step 3: The Long Burn

Now it's departure time (T), and you are ready to try that hour-long burn. First, you'll find that is not such a long burn any more, we've just taken 900 m/s off of it. But let's assume, for the sake of argument, that it's still an hour-long burn. (What are you doing, flying one of my"Megatankers?" :D ) typically, in KSP, your goal would be to center your burn on the node. So for an hour-long burn, you would start half an hour before the node.

However, in this case starting half an hour before your node, you're still going to be well on the wrong side of Kerbin. Burning then will destabilize your orbit and possibly send you into the planet. I think you at least need to wait until you're on the correct hemisphere of Kerbin (the side with your node). Personally, I like to wait until I can see around the planet. This means you will only be able to start 10 minutes or so before your node and your burn will run 30 or 40 (or 50) minutes late. That's fine.

Because 200 m/s into this burn you have Kerbin escape velocity, technically in solar orbit. Solar orbits are much more forgiving. The lines are much straighter and you're not moving nearly as many degrees around the sun as you would in the same time in orbit around a planet. You will find that the blue reticle wonders a bit toward the end of the burn. Follow it.

 

 

KDO4.jpg.

Forgive my enthusiasm, The orange arrow should be inside 80 km red maneuver node.

Because burning early will reduce the Pe altitude just a little. (That's why a 70 km Pe is too low.)

 

 

That's it. YouÃ're done. I do recommend a midcourse correction to refine your approach to your target planet.

Congratulations, you've mastered the Long Burn.

 


Optional Step 2B: Precise Departure Time

(to be inserted between Steps 2 and 3 above)

KDO is approximately a 10 hour orbital period, which means your departure time will be randomly up to five hours plus or minus of your desired departure time. I think that's good enough for KSP. But some readers in the thread below believe that is not accurate enough. (esp. for Moho) So I offer you this optional Step 2b, which will give you a precise departure time (if that's what you feel you need).

What we're going to do is turn your last 3 random, approximate, orbits into 2 very precise orbits.

 

Step 2B1: Wait until T minus 34 Hours (for Kerbin).

Given the KDO's 10 hour orbital period, approximately 34 hours prior to departure time, your ship will have just passed apogee and be descending toward perigee and you will have three orbits left. At this time, you should look to see what is your exact time of your next perigee. Now compare that with your desired departure time (subtract) and know the difference. (Rounding to the nearest minute is fine, but if you prefer seconds, go ahead.) It should be a number between 25 and 35 hours.

If it is 30 hours, stop now. You have nothing to do. Your departure will be on time. Otherwise do the following:

Step 2B2: Take that number and divide it in half.

 

Step 2B3: Now go to this orbital period calculator web page: https://dl.dropboxusercontent.com/u/1646976/KSP%20Calc2/index.html Enter that half as your desired orbital period. (click Total Seconds) Then enter your current perigee as the Pe (and click Calculate). It will provide an Ap that is higher than your current Ap.

Step 2B4: Plot and execute a maneuver at your upcoming perigee, prograde, to raise your apogee to that point.

That's it, in two more orbits you should be exactly at your perigee and exactly at your departure time.

Unless:

There is now a possibility the Mun is in your way. In that situation you have two choices: one, forget all of this and take your original departure time. Or two, take that original number you had and divide by 4 (instead of 2). Enter that number into the orbital period a calculator, and it will give you an apogee that is lower than your current apogee. That will be a retrograde burn (and a very tiny waste of fuel) and it will be 4 orbits, not 2, but it will take you to your departure window on time and without munar interference.

Or like I said you could forget all of this and just take five hours plus or minus.

 


But You Need More Time!

But what if your departure angle is less than 90°? Simple, just do the Mangalyaan Maneuver after you leave. :D

 

 

No, there are in fact two points in the orbit of a planet where the sun, the planet and your departure angle will line up, the other one is simply half an orbit earlier. Just subtract half the orbital period from your alignment date to find the other alignment date.

One HALF Solar Orbit (in Days):  
  Earth Time Kerbin Time
Kerbin 53.26 213.04
Moho 12.83 51.29
Eve 32.75 130.98
Duna 100.21 400.82
Dres 277.16 1108.64
Jool 605.68 2422.72
Eeloo 908.52 3634.08

However, when you do this, you have to reverse perigee and apogee. Take a look at these examples:

 

KDO5.jpg

 

A: Ejection Angle 86° to retrograde. (a common trajectory for Moho) So our nearest alignment date is one day after departure, T plus 1 day. (That won’t do.) We subtract 53 days and now have an earlier alignment date of 52 days before departure. But we reverse perigee and apogee, so our perigee point is away from the sun.

 

B: Ejection Angle 99° to prograde. Alignment date is T minus 2.5 days. But what if you want more lead time than that. (You really like lead time!) You can subtract 53 days and get an earlier alignment date of T minus 55.5 days. But this time our perigee point will be toward the sun.

 

C: Ejection Angle 15° to retrograde. (Does that go anywhere? Doesn't matter.) Nearest alignment date is 22 days after departure, subtract 53 days, and our earlier alignment date is 31 days before departure, perigee away from the sun.

 

 

Just remember anytime the planet is moving "retrograde-relative-to-T" (i.e. the opposite way from its direction on departure date) that's when you reverse perigee and apogee.

 


 

FAQ:

Q: What about ejection angles greater than 180°?

A: Doesn't happen. 180° from prograde is retrograde.

Q: Does this save fuel? (or waste fuel?)

A: Neither. The Mangalyaan maneuver is totally neutral. You’re doing the same burns and have the same total delta V, you're just doing some of it earlier.

That said, being able to find a precise interplanetary ejection angle can help you minimize delta V, that saves fuel. And this enables you to build ships with a lower thrust to weight ratio, less of the ship's mass is engines, and the ship can be smaller and lighter. That saves fuel, too.

Q: Are you a wizard?

A: No, just a KSP player who remembers his high school geometry.

Thanks for reading, hope it wasn't too long.

Again I'm neither a physicist nor a mathematician, but if you are, and have any input on this, I would love to hear from you. (Other comments are welcome too!)

Edited by Brainlord Mesomorph
UPDATED! Again!

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Q: Are you a wizard?

A: No, just a KSP player who remembers his high school geometry.

My high school geometry class wasn't enough for me to figure that out.

Thanks for reminding me I was ejukated in the lowest funded district in the lowest funded state in the lowest funded 1st world nation.

Great tutorial though. I'll try using this to remove the chemical engine escape stage from my ion probes.

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My high school geometry class wasn't enough for me to figure that out.

Thanks for reminding me I was ejukated in the lowest funded district in the lowest funded state in the lowest funded 1st world nation.

OK, high school geometry, and a lot of KSP, and several days in front of that white board!

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Q: Does this save fuel? (or waste fuel?)

A: Neither. The Mangalyaan maneuver is totally neutral. You’re doing the same burns and have the same total delta V, you’re just doing some of it earlier.

That said, being able to find a precise interplanetary ejection angle can help you minimize delta V, that saves fuel. And this enables you to build ships with a lower thrust to weight ratio, less of the ship’s mass is engines, and the ship can be smaller and lighter. That saves fuel, too.

As it happens, it DOES save fuel, because doing shorter burns near periapsis means you're closer to the highest-speed part of your orbit during a better percentage of the burn (see Oberth effect).

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As it happens, it DOES save fuel, because doing shorter burns near periapsis means you're closer to the highest-speed part of your orbit during a better percentage of the burn (see Oberth effect).

See, I told you I'm not a physicist! :D

Edited by Brainlord Mesomorph

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One critical thing you should really mention:

The first and foremost result any newcomer will very likely have when doing your "Mangalyaan Maneuver" is that they miss their departure window completely. That is because as you raise your apoapsis more and more, your orbital period increases. So when you want to do a burn at a specific hour, you might suddenly find yourself trapped in a very eccentric orbit for another two days until you get back to periapsis!

Of course, the advice to keep your apoapsis below munar encounter height is a sound one, because it not only keeps the Mun from throwing off your orbit, but it also keeps the orbital period relatively low. 8-10 hours or so at most? Still, chances are much bigger that you'll miss your departure window by 4-5 hours either way, than hitting it precisely. Unless, of course, you aim.

The last periapsis kick you perform should set your time to periapsis to a number that the time to transfer is divisible by. For example, if you have 47 hours left until transfer, eyeball something like 7 hours 50 minutes orbital period for your final parking orbit. Without using mods, the orbital period is equal to the time to periapsis when you just passed periapsis. Or twice the time to apoapsis as you're passing periapsis, if that's more convenient to read.

With this, your advice becomes sound even for things like RSS, where parking orbits like this take much, much longer to come around again. :)

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The first and foremost result any newcomer will very likely have when doing your "Mangalyaan Maneuver" is that they miss their departure window completely. That is because as you raise your apoapsis more and more, your orbital period increases.

Well, not "completely." (or it depends on your definition of "departure window." if you're trying to catch a moon as it rotates around a distant planet, I guess not.)

But on this webpage: http://alexmoon.github.io/ksp/ the dark blue splotch (minimum delta V) is a period of at least a day or so. Sometimes more. The orbital period of KDO (9000km x 80km) is about 10 hours. (5 hours either way IS my Launch Window) You may not get the exact minimum delta V, or the exact same time of flight, but the "first and foremost result any newcomer will likely have" is that he'll get there.

No one suggested a two-day orbit, and I don't know anyone trying to hit a one hour departure window.

BTW: its not "my Mangalyaan Maneuver" blame the Indian Space Research Organization :D

Edited by Brainlord Mesomorph

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I just felt it was worth mentioning. Missing a transfer window is distressing because an inexperienced player cannot tell what effect it will have and whether they should take the burn or not (and how to ajust for the imprecision). And considering that all it takes to explain is a single short paragraph, the guide just feels incomplete without it. Why give advice for a single specific case when you could give advice covering all possible cases instead?

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I'm outlining a simple method that works. (Have you tried it?)

The 10 hour orbit fits neatly into any reasonable interplanetary transfer window in the game. No one's going to miss a departure window following these instructions.

Sorry but this isn't "a critical thing I should really mention" at all. There is simply no need for this complexity that you are trying to add. And this isn't armchair speculation on my part, as it appears to be for you. I use this method, it works. (I'm sorry if "the guide just feels incomplete" to you. It doesn't to me.)

A better question is:

Why scare people by warning them of dire consequences that can only happen if they try to take this tried-and-true method to some ridiculous extreme?

Edited by Brainlord Mesomorph

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Why scare people by warning them of dire consequences that can only happen if they try to take this tried-and-true method to some ridiculous extreme?

Because they can, and quite a few will. If raising apoapsis is a good thing, raising it more is better, amirite?

I like to start some 30 Kerbin days before the transfer; stupidly raise apoapsis to about the Mun, then actually plan the last burn or two so that I will return to periapsis just in time. One needs to pay attention to the Mun, however, an unexpected encounter can ruin everything.

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This was pretty darn helpful, thanks! I tried to use MJ to get my huge "everything-but-the-kitchen-sink" spaceship to Duna, and was a little upset when it started the 40 minute atomic rocket burn by flying straight into the atmosphere. Thanks for doing the math for us.

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This was pretty darn helpful, thanks! I tried to use MJ to get my huge "everything-but-the-kitchen-sink" spaceship to Duna, and was a little upset when it started the 40 minute atomic rocket burn by flying straight into the atmosphere. Thanks for doing the math for us.

My ships have a kitchen sink! :D

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Hallo Mr. KSP Wizard!

I'm trying to figure out your equation in KSP time, 'cause Kerbin still goes round its' Sun once every Kerbal Year, so well... let's say converting kerbin time to earth time and back again is making *my brain* hurt!

In your equation I can't figure out where 3.38 comes from. It says 3.38° is how far Kerbin goes round its' sun in one day. But an earth year is 365 days and the total numbah of degrees is 360, so...... 360/365? To get degrees moved in one day? And a Kerbin Year is 426 (kerbin) days, so 360/426 to get degrees moved in one kerbin day? :confused:

I keep getting different answers, and feel kinda dumb I can't find where 3.38 came from. Or, maybe you really do just remember way more high school geometry than I do and I'm missing some piece...

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An Earth day is 24 hours which is 4 Kerbin days. You are correct that the Kerbin year is 426 Kerbin days, or 106.5 Earth days. For a Kerbin 6 hour day, 360 degrees divided by 426 Kerbin days equals 0.845 degrees per Kerbin day. But for an Earth day of 24 hours, 360 degrees divided by 106.5 equals 3.38 degrees per Earth day.

The part I'm trying to sort out is when the first burn occurs. I get the image, but how do I know when I'm there?

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Hello Mr. Explosions!

re: Kerbin time look at this web page:http://wiki.kerbalspaceprogram.com/wiki/Time

1 "Kerbin Day" (1 revolution of the planet) = 6 hours, 1 Earth Day = 24 hours (hours are constant)

1 "Kerbin Year" (1 Kerbin orbit around its Sun) is 106.52 Earth days. : 360 / 106.52 = 3.38

1 "Kerbin Year" (1 Kerbin orbit around its Sun) is 426.08 Kebin days. : 360 / 426.08 = 0.85

So in K-Time:

kD = (Ejection Angle -90) / 0.85

Where kD is the number of Kerbin Days prior to departure.

HTH

EDIT: That means Kerbals have a two-day Leap Year every 25 years! (OK now my brain hurts) :D

- - - Updated - - -

The part I'm trying to sort out is when the first burn occurs. I get the image, but how do I know when I'm there?

the first burn? aligning Pe and Ap w/ the sun? at (the new) Pe! in line w/ the Sun.

Edited by Brainlord Mesomorph

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EDIT: That means Kerbals have a two-day Leap Year every 25 years! (OK now my brain hurts) :D

If you go the Gregorian way, it's a one day leap year every 10 years, and a minus-one-day leap year every 100 years instead of the leap year which was due. In our calendar, the second cycle (every 100 years) is minus-one-day instead of Kerbin's minus-two-day-cycle; it just cancels with the leap day supposed to happen, so we get an excetional non-leap year (e.g. 1900). Then we (Earth) need an extra one day leap year every 400 years. So 2000 was not just a leap year, but an exception of an exception, and 2100 will not be leap year. Not many people realized that ;). There's still some round-up leftover, but it's so little that it disappears in the perturbations...

Thanks for awesome thread. Wish we could rate threads on all forums...

- - - Updated - - -

Eh, sorry for the kind of necro. I came here from your signature and didn't realize it was silent for a while...

- - - Updated - - -

Hey, we CAN rate on this forum!

- - - Updated - - -

Uh. My math is wrong somewhere. Damn. Will fix it! Fixed!

Edited by monstah

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Just got home from work and saw the pictures (firewall there...). They're very informative! I mean, I understood it from the text, but good pictures nonetheless. Easier for those who really don't want to think math (as for me, I really want to think math).

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OK Guys, I've updated the OP.

The technique has been refined with experience.

Tutorial has been rewritten with data added for ALL planets in the Solar System.

Optional Step 2B, provides for Exact Departure Time.

Game screenshot showing how to plot the maneuver.

Hope that helps,

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