Doozler

Is there an easy way to enhance that video? Seems like there must be a method of partially removing the ice/ water from the image as that bit is constant between frames, etc, etc.

Cheers everyone! Ah yes, this makes perfect sense. I think this didn't occur to me as I sttarting thinking about actualy landing first and then going sub-orbital. After that I realised I didn't have to actually actually land before going polar. I didn't then make the final step of burning direct from equitorial to polar sub orbital. The reason for the slightly odd change is that I was on extended science mission at the Mun, with a staion with MPL/ fuel in an equitorial orbit. I just needed the last couple of polar biomes and was on the margins for dV.

When doing an inclination change from an equatorial to a polar orbit, it is obviously more efficient to burn direct to your new orbit (i.e. at 135 degrees) than to stop completely, then burn south-north (by a factor of root 2). (and I understand it is even more efficient to raise Apoapsis and then change inclination there). But if you need to land at the poles, is it more efficient to stop all equatorial orbital velocity, then do a sub-orbital hop? It seems that this changes the situation as your polar "sub-orbit" does not need the full measure of dV. I have been doing the full inclination change and then a standard landing so far, but it occurs to me that this might not be the most efficient. Any thoughts?

Nerva: It's not "organic" per se, but "non-factory-farmed in the eightes" that's relevant. A lot of cows were fed ground up cow in those days. Organic is one way to ensure you're not eating "canibal cow"

I've had this a few times as well, on standard orbits. I switch to space center and back and that seems to help.

Here's a video of the "grasshopper" test rocket doing a sideways divert and back again. It almost looks fake! I wish I could land that well...

If you're trying to explain it to someone else, how about an analogy? If you have a room full of people and you start off with more people on one side that the other, then if they all push in random directions then they will eventually spread out - there's more push on one side of the room than the other. Analogies can be a bit dangerous, but an angry crowd is my go-to for gasses.

You should focus on the word "consistently" rather than the word "crash". Definitely not noobish! Ahem, "lithobrake", sorry.

Depends what you mean by "noob". In terms of the number of hours I put in, I've put in more on KSP than almost any other game - apart from maybe GTA. In terms of how far I've come since day one, I don't feel like a noob at all. In terms of how far I have left to go.. definitely a noob! SO badly it's scary! For the record, I started with 0.23. I have no kerbals out of Kerbin SOI, and I've only just landed my first rover on the Mun. No planes either. I read the forums a lot to pick up answers. Most things I want to know have already been posted, so I haven't needed to ask that many questions. When I do they get answered quick, so I leave again

I would say it's not all grind. Three biomes is easily done with three mun-and-back missions, but when you have a large number you start thinking a bit differently, e.g. I need a Mobile Lab, so I need a munar station, so I need to learn docking... Or, I make a mothership with many landers which means I need to build a big rocket... Sure, you CAN grind it, but I think it does have an aspect of "more is different" to it.

Yup, Zeno. You can keep burning for ever in the same sense that you can keep eating a cake for ever if you eat half, then 1/4, then 1/8.... but at the end of the day you only get to eat one cake (i.e. use one rocket's worth of dV)

I replied on the other thread, but was too slow! If you are talking about the moon stopping all motion relative to the earth (or both bodies stopping) then you can just consider this as a weird narrow "orbit" with a zero semi-minor axis, and an Apoapsis equal to the original height of the moon. So really your standard orbital equations should all still be valid. Edit: Ninja'd

The situation where both bodies suddenly stop can be thought or as just a "funny kind of orbit", where the semi-minor axis is zero. If you only stop the moon not the earth (relative to the fixed stars or something), then there is still an elliptical orbit as the system still has some angular momentum, you'll end up with a very narrow "orbit" with a small-but-not-zero semi-minor axis.

The difficulty is that, when falling, acceleration is not changing "constantly" in the sense of a=z*t, but changes depending on position (as described above). This is why you have to solve differential equations to find the answer. There's a derivation of escape velocity at Wikipedia which I think is the calculation you want, but it's not very illuminating: http://en.wikipedia.org/wiki/Escape_velocity#Derivation_using_G_and_M But now I have a question: If the "escape velocity" of a black hole is the speed of light, does that mean an object falling from infinity would reach lightspeed at the event horizon? That doesn't feel right to me but seems to follow from the definition of escape velocity.

Fair enough! Have you worked out the percentage of c that's needed to get your mass up to one ton? Edit: I think you're asking the question "what height should I drop something from to reach velocity v"? This isn't too hard - set potential energy equal to kinetic energy and solve for height. You won't be able to get enough speed from the earth or the sun. I suspect you'll need a super massive black hole....

For the above question, Yes. Start with constantly increasing acceleration a=z*t. Then integrate to get velocity, then integrate again to get position. Use the position equation to find the time of impact, and put that back into the velocity equation to get the velocity. I can supply more details (or the answer) if you want. However, I'm not sure that's what you want: I expect the teacher was not referring to relativistic effects! He was probably trying to say something like "a rapidly falling collation could do as much damage as a 1 ton weight" You could calculate this a few different ways, I'm not sure which is best. An easy way is "from what height would the collation have to fall to have a kinetic energy equal to the potential energy of 1 ton weight?" I doubt that's the right way to calculate ~"as much damage" but it's an easy place to start.

Yes, you are correct, in the thread I think most people are using specific impulse in terms of Mass, not weight, which is equal to the effective exhaust velocity. It's probably worth pointing out that KSP uses the weight definition, so to stick in KSP values to the above equations, you should replace Isp(mass) everywhere with Isp(weight)*9.8

Does low-quality content in wax crayon count?

First time I practice docking was with two purpose built craft - simple, cylindrical, perfectly placed RCS, docking port on the front, and it was relatively easy. Since then when I try it with "real" craft with fuel sloshing everywhere, solar panels at all angles, Kerbals wiggling about etc, it's been much harder. I'd agree with the suggestion to try using special "practice craft". I second the simple approach for the first few times (Remember to shift naval to target!): 1) Towards the green cross until velocity is 0, then 2) Towards the pink circle for a bit, wait until target is closer, then 3) repeat

It's a conversion factor, used to make sure that people don't get confused between units. Putting in the g (the 9.8m/s/s) means that you get the same number for Isp whether you measure it in m/s or feet/ s (or whatever). Criminally lazy people (like me) leave it out.

Exactly. So the real question is "what do you mean by outer space"? The other point is that although (eg) the atmosphere has in theory infinite extent, at some point it becomes "small enough" to ignore - and a sensible point might be "when the solar wind is more significant". Some sensible definitions of outer space might be: When you are closer to orbiting than flying (karman line) When you can complete an orbit without compensating for atmospheric drag When you can orbit for the lifetime of the mission/craft without compensating for atmospheric drag When the atmosphere is thinner than the solar wind When the Gravitational attraction from the body is no longer the largest (SOI) So for the ISS you would probably get YES, YES, NO, NO, NO Then you ask yourself, what do I care about? so for OP's question, what is important about the height of the ISS to you? Probably definition 3, in which case the answer is no. Rephrase the question more precisely and all will be clear

+1 here. Not needed the vast majority of the time, but would be easy to implement and has no real down-sides.

Yep, I'd like to see this. I think it'd be best further on in the tech tree, so you have to do a few "blind" landings on the Mun and Minmus. It should probably wait until after other planet biomes are added.. which will require a rebalance of the tech tree...etc, etc

Yup, that worked. I think I might have done that last time, it was appearing in the preview yesterday for some reason so I must have got lazy. Now where's that KSP Forum Latex Parser Petition for me to sign...

Gah, it was working yesterday Anyone have a solution?