Brainlord Mesomorph

A duration test of my jetboat/amphibious rover for Laythee. (I don't want to send colonists to a waterworld without a boat!) Straight off the runway heading east. Average speed 80 m/s, that's 180 mph! (or 155 knots - we are at sea!) Should this be a challenge? (is it one?)

Well, that's not the point. The point is getting a massive amounts of "payload" to where its going for a lot less "pay." Granted, long burns are not the most exciting part of my KSP experience, neither are the endless hours the VAB, or the countless laps back and forth to LKO. What is fun is solving the engineering problem. What is fun is arriving in the orbit of Jool with enough fuel will to do something when you get there. Like circumnavigating Laythee in a flying rover. Or several landings on Tylo. Take my MegaTanker: 360 tons of space craft powered by four nuclear thrusters. TWR of around 0.06. My guess is it will require an hour-long burn even after perigee kicking. But it arrives in the orbit of Jool with over 200 tons of fuel, and it costs $500,000. Including launching it to orbit. Now that’s fun! Also I have a method for long burns: I keep one eye on the screen while I putter about doing light housework. It makes me feel better about wasting a day playing KSP if I get the laundry or dishes done.

Which one has the most biomes? (the KSP version of which one NASA would pick.) I'm looking for a site for a permanent colony. (outpost really) - - - Updated - - - I'm planing on circumnavigation with fuel I brought from home. and I just finished designing my 80 m/s Jet Speedboat!

Not useless at all! I have burns of an hour or more. (See the tutorial in my sig,) All I need is enough thrust to blow out a birthday candle and enough fuel to do it for an hour and a half!

it's been done:

Bu...bu...bu.. But why?

It wasn't easy! (short controlled bursts, and patience) The spoke part is hollow, so the lab was pretty much the CoM. (after that I made a point to grab them from the end) Also: I learned that modules with their own RSC blocks were a lot easier to handle. Not autonomous, didn't even need their own monoprop, they would draw it from the constructor pod, But having RCS blocks made a big difference.

Sure! a small lander can with a tiny amount of LFO, a lot of monoprop, docking port at one end, and a grabber at the other: Docked at the Hub for refueling and some snacks. Chasing down a drifting module Bringing it home EDIT: wait... this thing is launched with the rest of the station parts. Does that make it 1 or 3? Another method I have used: two autonomous components, the station hub, and a scaffolding containing all the other modules. The hub goes and docks with a space station module taking it away from the scaffolding, then rotates around and docks with another module, etc, etc.

As a working hub it’s not very efficient. I tried it, I had this lovely science fiction painting in my mind, with my space station, and a wide variety of interplanetary tugs and landers, all arriving, and leaving, and working. In practice, it didn’t quite work that way. After using up the initial supplies, resupplying the space station was very problematic. Trying to send a huge tanker up to it, the tanker would consume huge amounts of fuel in orbital maneuvering and docking. Similarly, but to a lesser extent, thirsty interplanetary tugs would arrive in orbit, and run out of fuel and/or monoprop before docking. So I started using smaller “fuel ferries †to carry fuel from the space station to a larger ship. That is only a little better. In the end, I developed something the halfway between a fuel station, and a fuel ferry, I call the Fuel Tender, think of it as a single-use fuel station. It goes up, refuels as many ships as it can, and is then deorbited then recovered (it as parachutes). That said, I also have a very large orbital habitat currently in orbit.. It actually rotates for gravity!

Given that these are Kerbal "Atomic Motors" that somehow consume oxidizer. and that these are Kerbals, who can survive falls from great heights, don't need food, and have perfected some type of infinite rebreather technology, I figure, what doesn't kill 'em makes 'em stronger.

Ions are way too expensive for career mode. Its nukes for me! I even have a nuclear lander!

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?

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

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

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

http://forum.kerbalspaceprogram.com/threads/110213-Long-Burns-Interplanetary-Trajectories-with-Low-TRW-%28and-No-Math%29?p=1726429#post1726429

Just posted my tutorial on how to do exactly this! http://forum.kerbalspaceprogram.com/threads/110213-Long-Burns-Interplanetary-Trajectories-with-Low-TRW-%28and-No-Math%29?p=1726429#post1726429

Long Burns: Interplanetary Trajectories with Low TRW (and No Math) - or- How to Perform the Mangalyaan Maneuver under Visual Flight Rules - or- 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. 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. 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. 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. 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?" ) 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. . 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. 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: 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!)

I've said before I'll say it again: the launch pad is made of explodium!

(if I made it) Right before I put it on the launchpad.

Given "Revert to VAB" it seems unnecessary. (maybe if I was playing Ironman mode)

i call them "dollars." (just a crazy word i made up)

Bob's been stranded on Moho for a couple of years. and just I had to scrub his rescue mission. (But I did learn something new a interplanetary burns ) Waiting for the next launch window.....

"50 tons of fuel"

the RC module and the SAS module both have torque. if you move one to the other end of the ship it'll handle better.