tomf

You want to download the subassembly loader mod. It allows you to save part of an existing rocket and then load it later to attach to a different rocket. You can't do it with stock ahh ninja'd by someone who actually found the link.

my post here Calculating-interplanetary-delta-v Does a run through calculating the figures for anything without an atmosphere. For bodies with an atmosphere it is much harder.

Working out the perfect launch window is pretty hard, I looked into it and it seemed to come down to equations that weren't solvable algebraically. I think what people IRL do is draw up lambert pork chop diagrams and visually look for a minimum.

Another problem with Moho is that the eccentricity of its orbit means that the phase angle calculations done by protractor, mechjeb etc can be significantly wrong. At different points of its orbit moho can be traveling at very different speeds so it won't be where those tools based on circular orbits predict. I ended up setting an alarm for 3 days before the simple phase angle and then using a maneuver node to attempt to create a transfer. It was miles out so I waited for 12 hours and tried again, and again, and again. In the end the true windows was 6 days after the theoretical one. You want a transfer that just kisses the orbit of moho, and you want to do it at a time when moho will actually be at that point. If you come in to moho with any angle other than perfectly parallel to it's motion you end up with a massively more expensive capture burn.

For the complete mission you shouldn't sum the delta-vs like that, you should instead be looking at 4500ish of in atmosphere delta-v for you launch (so idealy you might use aerospikes) and then a thousand or so in vacuum (lv-n s would be perfect) for you transfer to duna, some more in atmosphere delta'v to lift of from duna (although the presure is low enough that each engine will be towards it vacuum isp) and then more vacuum delta v to get home

The optimum TWR is just above 2 during the more or less vertical ascent phase of the launch. When launching vertically there are two sources of drag on your rocket, gravity and air resistance. The gravity slows your craft by 9.8m/s (on kerbin's surface) every second you spend going up so to minimize gravity losses you want to go up as fast as possible. The air resistance slows your craft by an amount prportional to the square of your speed, so it is zero when you are barely moving and climbs rapidly. At terminal velocity (as given on the wiki etc) the two forces are equal and the combined drag is minimized - you aren't hovering burning fuel more than necessary and you aren't losing to much to the rapidly increasing drag. Since a TWR of 1 provides exactly enough force to overcome g and air drag = gravity you want TWR = 2 to keep you balanced at terminal velocity upwards. You want to be slightly over 2 because you still need to accelerate to to keep up with the increasing terminal velocity. Once you start your turn the required TWR decreases. I expect it is proportional to the cos of the turn angle but I can't prove that.

The quote on the website about doing free-fall experiments on a cat has made my day.

Your initial burn at kerbin is more but if you arrive at AP the difference in your speed and Moho's speed is larger than if you arrive at PE

It would depend on whether you want efficient or simple in terms of flight plan and number of parts. a compromise between the two would be If you aren't familiar with asparagus staging look under advanced rocket design on the wiki. Payload - orange + rcs fuel +solar panels + probe core + docking port = 38t Underneath = Orange tank and a mainsail Asparagus stagea around the outside 3 * (2 * mainsail + 56 units of rockomax tanks) That should give a delta-v of over 5000, you will probably have most of the final stage fuel left as well. Stick some sort of probe body on it and you can use the fuel to de-orbit the junk. If you want to maximize efficiency I would replace the central mainsail with 4* lv-45 and use aerospike clusters on the asparagus stages but the rocket will be a lot more complicated.

An orbit perfectly following the terminator would only work for a short time. The plane of the orbit won't rotate to match the rotation of Kerbin around the sun so in 1/4 of a kerbin year that orbit will be following the line of midnight/noon. In RL we take advantage of the oblateness of the earth to precess the orbital plane to keep it always aligned.

I have an eve 2-man sea level return lander, I originally wanted a 3 man version but the mass just exploded too quickly. If you land at one of the highest points of eve you should be able to do it though. My 2 man lander has a couple of vertical stacked stages and then a ring of clustered aerospike asparagus stages and a ring of mainsail asparagus stages.

Arriving at Pe is the most efficient. By the normal delta-v calculations it makes a difference of about 700m/s for the whole one-way journey. Be warned though that the eccentricity of Moho means that it isn't where the phase angle calculators expect it to be and with the inclination you might end up spending an extra 1000 or so getting your capture orbit. Don't do what I did and only allow 15% more fuel then the theoretical value and end up with two Kerbals stranded in Moho orbit unable to get home. There rescue mission is scheduled to depart in 8 days.

You can't have a sun-synchronous orbit in KSP. In real life they depend upon the precession caused by the oblateness of Earth. You could set up an orbit that follows the terminator now, but in 1/4 of a Kerbin year it will be aligned to orbit over noon/midnight.

Yeah Moho can be pretty evil because it goes around Kerbol so fast and has little gravity of its own. If you are just a few degrees off parallel to its orbit when you arrive, extremely likely due to it's inclination and eccentricity, then circularizing can cost several thousand more than you expected.

The delta v charts are reversible, except for landing and taking off on planets/moons with atmospheres. If it takes 950 to leave LKO, 100 to transfer to Duna and 600 at Duna to circularize then to get home from Duna is going to require 600 to leave Duna, 100 to transfer to Kerbin and 950 to circularize at Kerbin. (all these numbers assume a perfect Hohmann transfer and ignore the eccentricity and inclination). Of course you can aerobrake when you arrive at a planet so that delta-v is nearly free and getting back from Duna ends up costing less than getting there.

Astrophysicist, I recon I have spent more time in front of a kerbal spreadsheet (at work) than I have actually playing.

If you go straight to aerobraking on Laythe from you transfer orbit that bit will be almost free. For the return to Kerbin a simple plan involving leaving Laythe 70km orbit for a Joolian orbit and then burning again to go home will cost about 2200m/s. If you feel you can pull it of though you can get all the way to Kerbin with a single burn from a 70km Laythe orbit for an amazing 1062m/s. That Oberth effect is quite significant! Attempting to explain this to myself - a 70km Laythe orbit whips around pretty quickly at 200m/s. When you add another 1000m/s you leave Layth'e soi so quickly that it's gravity hardly has time to slow you down at all, you leave Laythe at 1200m/s + laythe's speed of 3223 which is enough to escape jool and take you home.

Birrhan - do you know the mass fraction of your rocket? I have a 2 man sea level Eve launch vehicle that comes in at 456 tonnes to launch a final in orbit mass of 2.25 tonnes. A total of 12072 delta-v. The start mass does include the parachutes, legs etc required for the landing which are ditched at the start of the ascent. I spent ages tweaking the staging and engine choice in excel so I would be impressed/annoyed if anyone can get that much delta-v with a significantly better payload fraction on eve. The ship is currently en route to eve, but I did test it with hyper edit so I am pretty confident that it will work. Whether the interplanetary stage has enough to get the Kerbals home again, we will have to wait and see.

When you are at you node you can use the other dimensions on a manoeuvre node to fine tune your trajectory. (or at any other time) If your issue is that you can't accurately see what effect your adjustments are having you can set a settign called something like CONICS_DRAW_MODE to 0 which will mean that you can zoom in on your destination body and see the resulting orbit drawn there while you adjust the node.

You can replace one ship in a persistance file with another ship fairly easily. What I would do is backup your main persistance file launch the fixed version of the ship (in a new sandbox if you don't want time passing, debris etc in your main game) open the new persistance file, find the relevant ship copy all the PARTs find the ship you want to replace copy the fixed ship's PARTs over the broken ones.

You can only have stationary orbits above the equator, basically you orbit at teh exact height where you go round the planet exactly once a day. As you are moving the same speed that the planet rotates, for someone standing on the ground you appear not to move. To pick a particular longitude you would go to an orbit slightly above/below the stationary orbit and when you are in the right place you move to the stationary orbit.

For planets without atmospheres I wrote up a calculation here http://forum.kerbalspaceprogram.com/showthread.php/27171-Calculating-interplanetary-delta-v

I'm curious about what factors make KSP harder and easier than their real life equivalents. I think the balance overall is pretty good and that most of these factors cancel out Its a game Building a rocket is done with a mouse not engineering, tools and materials The only failures you need to worry about are simple, structural failures and bad design quicksave and load currently no life support no room full of people monitoring every aspect of the mission. Physics things that make it easier Everything is small, the delta-v to leave Kerbin is smaller than earth, it takes less delta-v to visit all the bodies than RL equivalents Drag model doesn't penalize un-aerodynamic designs Free fuel pumping - makes asparagus designs practical Ion drives are more powerful than RL Physics things that make it harder The engines have slightly low ISP / don't use LH2 - 280s mainsail vs 363s space shuttle main engine in atmosphere Low thrust to weight ratio of engines - mainsail 27 vs f-1 75 Heavy tanks - ksp orange tank has wet/dry =9, Space shuttle orange tank = 28 heavy payloads - the apollo CM mass is pretty close to the commandpod -mk1-2 with rcs etc included - this isn't to scale with engine power drag model - I've not tried FAR but I hear that a more realistic drag model gives much lower drag for most craft. Does anyone have any other suggestions?

Lifting of from an airless world is easy, or at least calculating a minimum delta v is. You first need to build up enough speed that you would be doing a circular orbit with an altitude of 0, then continue burning to raise your AP, and another burn to raise pe as a normal Hohmann transfer. After our first burn we want an orbit with pe=2.5e5, ap=2.8e5 and sma = 2.65e5 At pe that orbit has a v of 1018m/s on the equator of Moho you are moving at 13m/s so our first burn is 1016. The Hohmann calculation will tell us the second burn is 27 So the total delta-v from the surface to a 30km orbit is 1043m/s The ideal landing is the same process in reverse so has the same delta-v WARNING - if you try taking of like this you are in danger of crashing into mountains, or even molehills. And landing like this will likely lead to Kerbals splattered all over the landscape when you don't judge it perfectly.

Doint the pork chop chart and working out the delta-vs for non-Hohmann transfers looks pretty hard, but calculating the numbers shown on charts isn't too bad. I typed it up but decided it deserved a thread of it's own. http://forum.kerbalspaceprogram.com/showthread.php/27171-Calculating-interplanetary-delta-v