Difference between revisions of "Talk:Kerbin"

Revision as of 03:26, 28 July 2015

Wrong Orbit display

I'm a newbie and don't really know how to edit that, but it looks like all the apoapsis/periapsis display are incorrect for a lot of celestial bodies. In the wiki, Kerbin have a 13 599 840 256 m apoapsis/periapsis while in game it's 13 338 240 256 m. Even if you add/substract 600 000 m it's incorrect. - Indeed 14:54, 23 December 2014 UTC

The value here includes Kerbol's radius (261 600 000 m), while the map view display does not UmbralRaptor (talk) 13:05, 23 December 2014 (CST)

Ok thanks :) — Preceding unsigned comment added by Indeed (talk • contribs) 21:26, 23 December 2014‎ (UTC)

Yeah this is what Note 1 references: “The distances are given from the bodies center not surface (unlike ingame)”. (And don't forget to sign your posts, more information is in the link) — xZise [talk] 16:10, 25 December 2014 (CST)

In-Game Descriptions

Should we really be correcting Squad's grammar? If we do then they aren't quotes anymore. None of the descriptions contain apostrophes even on words like "dont" so I don't mind adding them in but otherwise I strongly advocate against correcting grammar. The quotes should be quotes. Now since I only have one computer and the game won't function for me unless it is in full screen, I couldn't view both the in-game description and my transcription of it at the same time so I may have made some copping errors (though I did form a check on each) and these should be corrected. I don't mind going back cheeking and correcting the quotes though I won't be able to until later (probably in a few days) but I thought I would discuss my reasoning here first before I do. I don't want to get into an update war. - benschwab 17:36, 29 March 2014 UTC

Hi Ben! No, I agree with you; literal quotes must be literal. I could have made that clearer in my edit summaries. I meant I was correcting to match Squad's bad grammar. No war. But I have two computers and checked all the descriptions to be exactly as Squad has them in-game – Dres is particularly ungrammatical... --Brendan (talk) 19:50, 29 March 2014 (CDT)

Kay. Thank you. I must of have been auto editing while switching between windows and mis-read what you were doing. I appreciate you correcting my mistakes. - benschwab 5:02, 30 March 2014 UTC

Infobox

I've attempted to make an infobox template for planets, Template:Infobox/Body, to bring it closer in line to the stock parts here and planets on wikipedia. Any advice would be welcome. — Preceding unsigned comment added by Craigmt (talk • contribs)

• So far, so good. I think you have more columns than you originally anticipated, though, so many of the cellss are floating around. -- N3X15 (talk) 19:22, 24 September 2012 (UTC)
• Yeah, the length of the template is breaking some formatting stuff on articles it's transcluded in. It looks good, but I'd personally trim it down to mass/diameter/etc. and leave orbital information in the article proper. --  Trinexx  ►  19:34, 24 September 2012 (UTC)

So, leave orbital characteristics in the article itself and keep physical characteristics in the box? — Preceding unsigned comment added by Craigmt (talk • contribs)

• That's my personal recommendation, at least. I imagine other people have their own opinions on the matter. --  Trinexx  ►  20:40, 24 September 2012 (UTC)

Links

All the links used in the references are broken. Just an FYI. -- N3X15 (talk) 09:55, 1 October 2012 (UTC)

'Kermunist'

Should this term really be used? Thecoshman (talk) 18:28, 7 October 2012 (UTC)

There are no factions in the game, so it would be better to not have that in there, as it'd cause confusion. We do need a Manual of Style to cover stuff like this. -- N3X15 (talk) 20:30, 7 October 2012 (UTC)

Atmosphere

I edited the atmosphere section to make clear the units of d_k & p_k after working through gamedata. Still unsure what units p_k are in FlightGlobals.getStaticPressure() values are in atm, with 1 atm = 101325 Pa to get an absolute static pressure value needed for ideal gas law

-Kellven

Should it be mentioned that because Kerbin has plant and animal life, that therefore the atmosphere is similair to our own? That would be a bit quick on the draw given that even our atmosphere is extremly hostile to some earthbound organisms. --Azivegu (talk) 08:33, 13 December 2012 (UTC)

Orbital information

it mentions in the infobox that Kerbin has a o ecc and 0 inc orbit.. i don't think thats correct, can someone plug the correct figures in?

Maltesh's rip of the 0.18.2 orbital parameters has those, and a quick check in map view shows Kerbin's altitude and speed not changing anywhere in the orbit, so it should still be correct. UmbralRaptor (talk) 06:48, 1 February 2013 (CST)

I'm not sure it's the best place for this information, but I linked to the planetary data spreadsheet and text files at Celestials#Notes. — Elembis (talk) 10:54, 1 February 2013 (CST)

Well, that will teach me to open my big mouth before checking my facts. I guess the inclination of the other bodies is measured relative to Kerbin. Oops. Kahlzun (talk) 21:14, 8 February 2013 (CST)

Why is sidereal orbital period showed in Earth days (106 days 12 hours 32 min)? That has no use, it obviously should be in Kerbin days: 426 days 32 min. I have Kerbin calendar and want to know how many days it takes to Kerbin to cover X radians. --Masuk (talk) 09:59, 28 January 2015 (CST)

It was from before Squad redefined 1 Kerbin day to 6 hours. Before that all calculations used 24 hours. And “obviously” that isn't obviously as it's not visible which day length is chosen (Kerbin's sidereal day, Kerbin's solar day, Earth's sidereal day or Earth's solar day). It seems obvious for Kerbin, but not for the other celestial bodies. — xZise [talk] 10:43, 28 January 2015 (CST)

Sidereal day length changed

With the release of 0.24 the solar day is now exactly 6 hours long. Is my understanding correct? What would make the sidereal day then? Is there an objection to updating the page to reflect this new information? Benschwab (talk) 20:25, 24 July 2014 (UTC)

I'm not sure if that hadn't already changed before. — xZise [talk] 17:14, 25 July 2014 (CDT)

Okay thanks to User:UmbralRaptor on the IRC I was able to make calculations. He placed a command pod on top of the launch pad at a altitude of 72 m in 0.19, 0.23.5 and 0.24.2 and always gets a orbital velocity of 174.6 m/s.

The orbital velocity while standing on the surface is basically the rotational speed at that altitude. Now using ${\displaystyle v={\frac {s}{t}}}$ we can calculate the time required to do a full circle with a radius of 600 072 m: ${\displaystyle t={\frac {2\cdot \pi \cdot r}{v}}}$ = = 21594 seconds (or 5h 59m 54s) which is almost 6 hours. When you do the reverse you get 174 554 mm/s which rounded gets you 174.6 m/s. This makes me already pretty confident that the sidereal rotation is 6 hours and not the solar day.

Now lets give it a chance and use 6 hours as a solar day. But to calculate the rotational velocity the value needs to be converted into the sidereal rotation period, because on the way how the calculation works: When viewed from the north (or south) pole the craft on the launch pad is doing a circle and the circumference depends on the radius. To calculate a constant speed (a little physics: the velocity vector changes because the craft "curves" but the magnitude stays the same which is afaik called speed) the distance has to be divided by the time it took to cover that distance. The distance, is the circumference and the time it took that distance is the sidereal rotation period. If you use the solar day for that time you would need a larger distance (in this case, in theory the solar day can be shorter than the sidereal day) because Kerbin does rotate a bit further.

To convert that I used the formula from Wikipedia and plugged in the solar day length and sidereal orbit: 21459 seconds (or 5h 59m 9s) (for validity of the formula, c is the orbital period, x is the length of the solar day and y is the length of the sidereal day)

Using that we get 174 964 mm/s which is rounded 175 m/s. But that speed is considerably larger compared the previous result.

We also checked if the orbit has changed (which would have changed the length of a sidereal day if the solar day is fixed at 6 hours and got 13338240256m which is 261600000m lower than the value shown here in the infobox (13599840256 m) but is exactly the radius of Kerbol. And the altitude in KSP is given from the body's surface so that checks out. But even though the orbit must have been considerably larger to get solar day and sidereal day close (and I'm currently not in the mood to determine how large the orbit must be that both length of days are close enough that solar day length would make sense).

So unless we didn't consider a specific factor I gather that the length of the sidereal day hasn't changed contrary to the changelog.

Also if you think about it, what would be the advantage. Defining times based on their sidereal values makes it easier because you don't have to take the orbit into account. I also revert the change in the infobox to a sidereal day length of exactly 6 hours. — xZise [talk] 22:22, 4 August 2014 (CDT)

Map

Does anyone now what projection the map is in? a link to the relevant page would be nice, i am refering to the one in the Topography section

What map do you mean? — xZise [talk] 02:37, 22 April 2013 (CDT)

I would guess Equirectangular projection as the one degree in both axes in File:Kerbin heightmap.jpg have always the same length. — xZise [talk] 08:15, 18 May 2013 (CDT)

Orbits

I crunched the numbers using a few different methods and I keep getting a discrepancy between my numbers and the listed figures (I calculate a GEO orbital velocity of ≈1109 m/s, a difference of exactly 100 m/s). Can someone confirm this? Otherwise, tell me why I'm wrong. ;) -- Eurousalas (talk) 14:51, 22 April 2013 (EST)

I can confirm both. This confirm your value while this confirm the value of this page. What formula do you use?

${\displaystyle v={\sqrt {\frac {G\cdot m}{\sqrt[{3}]{\frac {p^{2}\cdot G\cdot m}{4\cdot \pi ^{2}}}}}}}$

m = orbited body mass, p = orbital period

— xZise [talk] 14:54, 22 April 2013 (CDT)

Thanks to a friend, I know what your mistake is: The GSO altitude is given relative to the planet's surface not gravity center. Now {{OrbitVelocity}} and wolfram alpha need this value to be from the gravity center. So you need to add 600 km to your altitude and then you get the correct value. — xZise [talk] 15:03, 22 April 2013 (CDT)

Ack, I knew it would be an obvious mistake. Thanks for pointing that out, it would have bugged me for a long time. -- Eurousalas (talk) 17:46, 27 April 2013 (EST)

Map

I'd like to note that both true-colour maps are outdated. Saik0 on the KSP forums has made maps which are not outdated and in Full-HD and uploaded them on http://www.kerbalmaps.com/ . The website includes anomaly locations (optional), elevation maps, slope maps and true colour maps for all bodies in the Kerbol system except for Kerbol and Jool. He also put those maps online for download at https://mega.co.nz/#F!8EAnSL5Z!OAyLBQqrQtlmfL0GT8fcQg!EBYyXRxA . I have not uploaded any of these images myself yet, because e.g. the Duna true-colour map is something like 77.5 MB where 2 MB are allowed and I'm unable to compress them. I came here to ask if someone would be interested in compressing these for the community. Oh also, here have his KSP forums thread: http://forum.kerbalspaceprogram.com/showthread.php/25148-Web-maps-of-all-bodies-(0-4-ALL-bodies-mapped!)?p=307335#post307335 MmPMSFmM (talk) 07:42, 2 July 2013 (CDT)

I'm wondering where the Kerbin map is on your download link? Here are my suggestions to compress them:

• Use a compression format like png/jpg
• Reduce the resolution of the image (may not be needed when compressing with jpg)

— xZise [talk] 08:36, 2 July 2013 (CDT)

I just found out why you can't find the Kerbin files on that MEGA site: on the forums the author says that "it is his first map and only has half the resolution of the others" and that he will upload it once he redoes it. Also I'm having problems with that MEGA site and Firefox, can't download the files, because of some message about accepting another message that doesn't appear for me... --dgelessus ^{(talk · contribs)} 09:23, 2 July 2013 (CDT)

I'm able to download the images. I tested Gilly but when I tried to edit gilly_0.19.1_0.02197265625_-180_90_180_-90_16384x8192.elevation.tif (which I guess is the most important type of map?) I can't open or edit it because: convert: Sorry, can not handle images with 32-bit samples.. Maybe I find a way to fix this and I'll upload some files. — xZise [talk] 19:10, 2 July 2013 (CDT)

Making Geosynchronous Orbit More Exact?

I see the formula for the calculation of geosynchronous orbit, but that only gives the orbit within the tens of metres. Through some calculations I've found that the closest I can get is 2868750.72505 metres. Would we be willing to change it?

Considering that it is practically impossible to get the orbit that exact, I don't think it's necessary. 2868.75km is more than close enough.--Ruedii (talk) 19:13, 23 July 2013 (CDT)

Elevation update

With topology changes in 0.21, the highest known elevation will have likely changed. My current efforts with a 52 million sample file from ISA Mapsat show a new highest elevation of 6765.1 at location +61.6,+46.34, but I want to confirm that with a landing before updating the page and also wasn't sure if I had missed any higher points.--Khyron (talk) 02:12, 28 July 2013 (EDT)

Surface gravity

I suggest adding a percentage value to the surface gravity inside the info boxes of the planets. This percentage value, denotes a comparison to Kerbin. So the 9,81m/s² acceleration of is 100%, and the value of other planet is an according value. Another way of denoting this value, would be to set Kerbin's s as 1 and then use that to compare the gravitational acceleration of other planet. Polemon (talk) 14:39, 11 September 2013 (CDT)

You are talking about using “g-force”: 1 g = 9.81 m/s². This is in theory possible, but I'm not quite sure if this is really helpful. The template is already enormous and Kerbol System/Table for example show them for you. — xZise [talk] 15:06, 11 September 2013 (CDT)

Why useful: When designing crafts that are supposed to land on other celestial bodies, I like to test them on Kerbin beforehand. Doing this kind of calculation is "OK-ish", but I don't see why it shouldn't be in the infobox as well. Polemon (talk) 15:12, 13 September 2013 (CDT)

btw, I fixed your comment - looks like you accidentally overwrote xZise's message. --dgelessus ^{(talk · contribs)} 15:21, 13 September 2013 (CDT)

Okay as there is already enough space, I added it. — xZise [talk] 05:04, 15 September 2013 (CDT)

G factor

Does g = 9.82 m/s² ? Specific_impulse#Conversion_factor discusses back-calculating g, and arrives at a value closer to 9.82. Any thoughts on breaking this out in to its own page, or at least "correcting" it here?--BlobKerman (talk) 07:50, 26 May 2014 (CDT)

The gravitational acceleration at the surface shown in the infobox is calculated, based on the mass and radius of Kerbin and using {{Body data/Gravity}}. So I wouldn't redefine g to 9.82 m/s². The conversion factor is arbitrarily chosen and "coincidentally" the same as the average surface gravity for real-world rockets. You could use 1 m/s² and in the US they would then choose something like 3.3 ft/s² and would get the same results. — xZise [talk] 13:08, 26 May 2014 (CDT)

At least when I last checked, g for Isp and g for Kerbin's surface did not match up. Hence 9.82 for engine efficiency, but 9.81 for weight. UmbralRaptor (talk) 14:31, 26 May 2014 (CDT)

KSC Biomes.

I've updated the KSC Biome section, I believed there were 11, but reports on the forums are showing that there could be as many as 30? Can anyone verify this? I'll try and take another look later and do a bit more exploration, but I only found 11 on the ground with EVA - Science Alert going off every time I stepped into a new biome. — Preceding unsigned comment added by Severedsolo (talk • contribs) 10:30, 18 October 2014‎ (UTC)

With 0.25 all the buildings at KSP are distinct biomes. HenrikOOlsen (talk) 02:52, 2 December 2014 (CST)

Atmosphere: altitude vs. pressure

So yesterday I dropped a rescue pod into kerbin and while I waited it to land I kept the barometer display visible and took some readings. I post it here for use of the community.

Nando.sm (talk) 01:02, 27 October 2014 (CDT)

It would be interesting to have the corresponding calculated pressure/specific impulse in each row. (although I'm pretty certain those numbers are matching closely) — xZise [talk] 05:37, 27 October 2014 (CDT)

Atmosphere composition

In the atmospheric composition was added. How could that be determined? — xZise [talk] 18:52, 17 January 2015 (CST)

The gas molecular weight is calculated from the ideal gas law. The known sea level conditions are pressure (101325 Pa), temperature (293.15 K), and density (1.2230948554874 kg/m^3). From this it is a straightforward computation to derive the average gas molecular weight of 29.42 kg/kmol. We know oxygen (M = 32) is present, so for the molecular weight to be 29.42, the balance of the atmosphere must be a gas lighter than M = 29.42. The only candidate gas that can reasonably exist in large abundance is nitrogen (M = 28). Hydrogen, helium and neon are too light and would escape, methane and ammonia are either broken down or removed through natural processes, and carbon monoxide would oxidize to CO2 and be dissolved into the oceans. That leaves just nitrogen, though other gases are likely to exist in small quantity as in Earth's atmosphere (all gases other than nitrogen and oxygen make up just 1% of Earth's atmosphere). Assuming a nitrogen-oxygen atmosphere, the two gases must exist in the proportion needed to yield an average molecular weight of 29.42 kg/mol. This occurs when the percentages are 65% nitrogen and 35% oxygen by volume. — OhioBob (talk) 17:53, 19 January 2015 (UTC)

Okay that makes mostly sense except that we don't know much about Kerbin's flora and fauna and if it works like on Earth. But using that assumption anything could be possible. — xZise [talk] 05:29, 20 January 2015 (CST)

I've edited the article to make it more clear what the assumptions are. You are right that there are things about Kerbin that we don't know, but it seems to be the game's intent to make Kerbin Earth-like. If we to go completely sci-fi we can make the atmosphere anything we want; however, if stay with what is scientifically plausible, then the best explanation is a nitrogen-oxygen atmosphere. I've also done some calculations and found that a real-life planet of Kerbin's size wouldn't be able to retain an atmosphere. — OhioBob (talk) 22:48, 20 January 2015 (CST)

Have you ever heard of reaction kinetics? Have you ever warned not to deal the oxygen containers with oiled hands? Do you know what would happen at 35% concentration of oxygen? Almost everything would catch fire immediately (even the nitrogen when a rocket engines starts)! Why not mix Argon (M=40) in your equation? It is dense, inert and even earth atmosphere contains copious amount of it! NWM (talk) 01:35, 21 January 2015 (CST)

With three gases it's impossible to calculate the proportions unless one is known; however, adding a heavy gas like argon could certainly bring the oxygen percentage down to something more Earthlike (O2 decreases by about 3% for every 1% of Ar added). I'd certainly be willing to remove the sentence from the article that talks about 65/35 N2/O2 since that might be too big of an assumption to make. However, the part about the molecular weight being 29.42 is entirely valid, as that is computed directly from given data. I also think the part about nitrogen being the other major constituent is a reasonable conclusion. What do you guys think? — OhioBob (talk) 02:14, 21 January 2015 (CST)

I've removed the part about 65% N2 and 35% O2. Instead I simply said that nitrogen is the "majority constituent", which is almost assuredly true given the molecular weight. — OhioBob (talk) 11:25, 21 January 2015 (CST)

Also note that the peak oxygen level on Earth was about 35% in the Carboniferous period, so that level is not unprecedented or impossible. — OhioBob (talk) 13:45, 21 January 2015 (CST)

Atmosphere — version 1.0 changes

I've updated the atmosphere section to include changes made with the release of v1.0. Much of what was previously written in no longer applicable. For instance, pressure is now evaluated using a floatCurve, thus the old isothermal/constant scale height method no longer works. Atmospheric properties are contained within a configuration file named CelestialBodies.cfg, to which I have access. That facts and figures I've provided are those used within the game. I've also have a couple of images that I want to include that show a graph of temperature and pressure vs. altitude, though I keep getting an error when trying to upload. This may be related to the migration problems that the forum is having. I also couldn't get the formula text to work, which may be related to the same problem. After the migration issue is resolved I hope to add the images and correct the math text. — OhioBob (talk) 17:20, 20 May 2015 (UTC)

Terminal Velocity

I removed the section about terminal velocity because little of what was written is applicable to the new aero model. A table of terminal velocity is meaningless because there are now variable factors that go into it. If anybody wants to rewrite that section, the old text is still there; it has just been bracketed with comment tags so that it doesn't appear in the article. — OhioBob (talk) 17:20, 20 May 2015 (UTC)

Well there is a version history so they could just use the text from there. — xZise [talk] 16:45, 21 May 2015 (UTC)

OK then, I've deleted the text. — OhioBob (talk) 14:46, 22 May 2015 (UTC)

4500 m/s is obviosly wrong now. Should be changed to 3700-4000 m/s (dependent on ascending pofile and aerodynamics) — Preceding unsigned comment added by Ksbes (talk • contribs) 19:55, 1 June 2015‎

Sidereal Rotation Period

Although the Kerbin calendar day is 21600 seconds (6 hours), I just discovered that the actual sidereal period is slightly less. I observed the same star on the eastern horizon on the first day of each year and found that it was rising sooner each successive year. I warped ahead 20 years and found that it rose a total of 748 seconds earlier, or 37.4 seconds each year. This means that the time it takes the planet to rotate 426 times is 37.4 seconds less than a full year. Therefore, the actual sidereal period is (426*21600-37.4)/426 = 21599.9122 seconds. This makes the solar day 21650.7246 seconds. We should also find that the Sun rises sooner each successive year by the amount, 426*21600-425*21650.7246 = 42 seconds. Observation of the rising Sun has proven this to be true - the Sun rises 42 seconds earlier from one year to the next on the same calendar day. OhioBob (talk) 03:26, 28 July 2015 (UTC)