Brainlord Mesomorph

I have a standard 50-Ton lifter for most payloads. (i try to keep a lander and fuel under 50 tons) If its much heavier or lighter then "yes."

but the TWR has changed! (I'm slowly getting this)

They seem to have fixed that in 0.25 - I tested it, I hung a weight from one those from a gantry - try it in 0.24 I had to build them around scaffold. the new ones look much better EDIT: I also built a spaceplane that involves a MK2 Two-way adapter connected to a two-way to large tank adapter - then to a large tank. it works too.

the Rockomax 55 has as Isp of 320, if I use 3 of them is my total Isp 1280? or still 320?

So i *don't* swap that out for the gravity of Tylo? (When do calculations for Tylo) ---- (light slowing dawning) OOOOHHHHHHH, so I can use either feet or meters -- if I put in G as feet per second^2, I get deltaV in ft/s --- (DAMN this has got to be THE most educational game I ever played)

I've heard of the Rocket Equation (Robert Goddard, right?). I get this: [TABLE=width: 566] [TR] [TD][/TD] [TD]Isp[/TD] [TD]WetWgt[/TD] [TD]DryWgt[/TD] [TD]G (Kn)[/TD] [TD]DeltaV (Kn)[/TD] [/TR] [TR] [TD]Gamma II [/TD] [TD]320[/TD] [TD]32[/TD] [TD]18[/TD] [TD]9.81[/TD] [TD]1806.183124[/TD] [/TR] [/TABLE] But now I'm confused. That is the max Delta V for a rocket launching from the surface of a body with gravity "G" right? But what about in space?

totally reused. Lander CM and return lander.\ you'd be surprised what a low CoG that has. Its down where the landing gear is. ( the thing on top is light, and I use a drop tank during decent so it lands full) will that work as a formula in XL, or not? (I was just trying that)

math? --- sigh --- (thanks, I will add that to my KSP excel workbook)

restageing is perfectly comulent.

OK wrong calculator. Looking for a delta V calculator to tell me if this thing has enough engines and fuel to make orbit of Tylo. its basically a trash can; the big gray fuel tank, a two-man lander can, and three Rockmax 55 radials, plus gear and a tri-sci module adds up to 32 tons. With the aid of a drop-tank, can do a Munar Mission (from LKO) by itself.

Trying to figure out if my latest lander has enough delta V fro Tylo. According to this web site:http://iyates.co.uk/ThrustCalc/ It has enough delta V for escape of Kerbin! But it does not. Suborbital apogee of 48 km. Is the website wrong? Or is it just the atmosphere of Kerbin?

I know what launch clamps do. My point is you only seem to need them if you add S1-SRBs.

Building a huge liquid fuel rocket. Stands on the launchpad fine. add 2 S1 SRBs (not even near the ground) and KABOOM! remove the SRBs - all's quiet put them back - KABOOM! add launch clamps to the SRBs, ... no kaboom. HTH

I had a very scary near miss with some orbit debris the other day. I'm docking two ships and one of those little grey crosshairs out in the distance is getting closer, and closer, and closer ... within a kilometer... It was stage separator ring. It came so close it stopped giving me distance and actually cast a shadow across the other ships as it passed. had to be within 10 or 20 meters. Scared the hell out me. (one of the things I love about this game.)

if anyone's watching this thread, I've continued it this thread: http://forum.kerbalspaceprogram.com/threads/97857-The-Long-Burn-Revisited-Departure-Orbit-Achieved%21

Not a physicist. I didn't know what Isp was (till now) but I did realize that thrust divided by fuel consumption would be a thing. So, thank you, I'll accept your congratulations. (definitely not the first time that this has happened to me. I am apparently more creative and intelligent, than I am educated.)

I divide the Total Thrust (in kiloNewtons) by the fuel consumption (in tons per second, vacc or asl). The resulting number is good for a comparison. And I think means : kiloNewton-seconds per ton of fuel.

The 4 LV-909s cost twice what one poodle costs.

Well, if the original problem was a 90° change in direction in 15 minutes in LKO, in KDO you have a 90° change in direction in 2 ½ hours. Is that our upper limit? Probably not. Because 200 m/s into the burn you already have escape velocity, you are interplanetary, and your trajectory is basically a straight line, so there may not be an upper limit. Torch ships anyone? One thing I forgot to mention: The departure burn starts at perigee regardless of where your final node is (that is, for a long burn). In my example, my Jool departure node was exactly where it is on that guys chart. The burn was 46 minutes, so I should’ve started 23 minutes early. But that would have been 10 minutes before perigee and wouldn’t have worked. So I started to burn precisely at perigee, and it ran 10 minutes long, but like I said by that point your course is basically a straight line so it doesn’t matter.

is exactly what I'm saying... Yes, and all of those departure angles will work with a basic prograde or retrograde Departure Orbit. (maybe not the absolute minimum delta V, but it does work) and what you just called the 26 day 1/2 rule just shows you what will be prograde or retrograde in 26 days. (as a matter of fact that's exactly where my Jool burn was)

The Long Burn Revisited : Departure Orbit Achieved - or - “It’s not the size of your thruster, it’s how you use it.†by Brainlord Mesomorph A 70 ton rocket delivered on a direct Kerbin/Jool transfer using only one nuclear thruster. Final departure burn was 1700 m/s and took 45 minutes. Departure altitude: 100km. No staging (after orbit). Burned 30 tons of fuel. All stock, no mods. (except Alarm Clock). Continuing our conversation from this thread: http://forum.kerbalspaceprogram.com/threads/97459-Help-with-plotting-interplanetary-trajectories We were discussing the problem of large underpowered ships being unable to achieve interplanetary transfer orbits from LKO. The issue being that the planet Kerbin is so small and LKO is so tight, that within 15 minutes you’re changing direction 90°. So if your ship can’t complete its departure burn within 10 or 15 minutes, you just can’t do it. It was explained that NASA’s solution to this problem would be a series of short perigee burns carefully placed to extend the apogee in the direction of the spacecraft’s final departure all the way up to SOI, creating a highly elliptical orbit the Earth at one end. Then one final burn at perigee to send the craft on its way. The problems (we thought) this created in KSP were twofold; first, we don’t know our departure angle until we’re in our launch window and we’ve actually plotted our course. And two, building up the orbit all the way to SOI would take too long and the final orbit would take almost a month and you’ve missed you launch window. One reader, metaphor, pointed out you didn’t need to get apogee all the way up to the edge of the SOI, just within lunar orbit would be enough. And that you would have time to do that within your launch window. And he was right. An apogee of 9,000 km is a 10 hour orbit. And you do have time to plot a course to your destination, use that node as perigee, build up your departure orbit, plot your final burn to your destination, and execute it, for one ship. If you hurry. But you don’t have to. Because our second misconception was that you don’t know the angle of the departure burn. Of course we do! This isn’t rocket science, OK actually it’s precisely rocket science, but this part is easy: If you’re using a launch window calculator to find the launch window with a minimum delta V, then your final departure angle will always be basically prograde or retrograde to Kerbin’s orbital track. Indeed, that’s why it’s minimum delta V. (right?) This means our departure orbit simply must be parallel to Kerbin’s orbit at the time of departure. With apogee prograde of Kerbin if you’re going to the outer solar system, and retro grade if you’re going in. How does this information help us? Well, Kerbin’s orbital period is approximately 106 Earth days. This means at 26 ½ earth days before your departure window, your departure orbit points directly at the sun. All you have to do is,26 ½ earth days before your departure window, line up your perigee and apogee points with the sun. You don’t even need to establish the departure orbit yet, you just need to mark your perigee and apogee points in space. Then you have 26 ½ lengthy earth days to build up your departure orbit, and once it’s marked in space, you can park other ships in it! You can park 10 ships 1 hour apart in this orbit weeks ahead of departure, and when your departure window opens you’ll have a full 10 hours to plot all 10 departure burns, and a full hour for each one, secure the knowledge that nothing will overlap. And that you’re using minimum delta V for everything. I want thank everyone who helped me figure this out. I hope I have paid the community back by telling you about my 26 ½ Days-Ahead-Point-Your-Orbit-At-The-Sun Trick. Should we have a new official abbreviation? KDO: Kerbin Departure Orbit. Defined as an orbit with a perigee of 100 km, an apogee of 9,000 km, parallel to Kerbin orbit at the time of departure.

we talked about that, it would take too long the last orbit takes a month and you miss your window. FYI

Current mission is Moho (no atmosphere) but you're right about Jool.

EXACTLY! and I think I have a method for doing it in the stock game! (The trick being that you only need to know the current node and the next node at any given moment, and you have PLENTY of time during one burn to calculate the node after that.) I have some experiments to run in sandbox mode. (God, I love this game!) I may have found a few feet on untrodden snow here, I may get to write my own tutorial "Mastering the Long Burn" I'll get back to you guys in a couple of days. THANKS! (kOS exists? I thought that was just a joke on a screen in the new cockpit)

But that won't solve the problem that 10 minutes in to the burn the ship is pointing (and moving) in the wrong direction. it will just do that in five minutes.