Difference between revisions of "User talk:XZise/Atmospheric Height"
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:: I agree that stopping the atmosphere at 10⁻⁶ atm everywhere would make more sense. As for the simpler computations, I was referring to orbits above the threshold, for which we have closed formula for pretty much everything as long as we ignore drag. Indeed, having a relative threshold instead of an absolute one for the top of the atmosphere does not simplify anything at all. So you can consider that KSP is at fault there, even if that is not really “wrong”, just a poor choice. — [[User:Yoha|Yoha]] ([[User talk:Yoha|talk]]) 03:48, 27 December 2014 (CST) | :: I agree that stopping the atmosphere at 10⁻⁶ atm everywhere would make more sense. As for the simpler computations, I was referring to orbits above the threshold, for which we have closed formula for pretty much everything as long as we ignore drag. Indeed, having a relative threshold instead of an absolute one for the top of the atmosphere does not simplify anything at all. So you can consider that KSP is at fault there, even if that is not really “wrong”, just a poor choice. — [[User:Yoha|Yoha]] ([[User talk:Yoha|talk]]) 03:48, 27 December 2014 (CST) | ||
+ | ::: Well I think it's a bug because it looks like someone just forgot the division. And of course it can't be ”wrong”, because maybe physics in KSP work differently which would explain the different “border”-pressures. It is “just” an inconsistency which doesn't make sense. — [[User:XZise|xZise]] <small>[[[User talk:XZise|talk]]]</small> 13:09, 27 December 2014 (CST) | ||
“you know could construct an example with an sea level pressure of 10⁺⁶ atm resulting in a pressure of 1 atm at the “atmospheric height” which is the pressure at Kerbin's sea level.” | “you know could construct an example with an sea level pressure of 10⁺⁶ atm resulting in a pressure of 1 atm at the “atmospheric height” which is the pressure at Kerbin's sea level.” |
Latest revision as of 19:09, 27 December 2014
“So applying Jool's “atmospheric height pressure” on Kerbin means, that you can savely orbit at about 56 km which is way lower than 69 km.“
In KSP, pressure is discontinuous at atmospheric height: when the wiki says that the pressure is 10⁻⁶ atm at atmospheric height, it means that it is 10⁻⁶ atm right below and exactly 0 atm (i.e. no drag simulated) right above. Orbits on Kerbin above 70 km and orbits on Jool above 139 km are stable because pressure is exactly 0 atm.
An orbit on Kerbin at altitude 69,077.552 m (1 mm before top of atmosphere) would experience 10⁻⁶ atm while an orbit on Jool at altitude 138,155.10 m would experience 15×10⁻⁶ atm. Both would eventually decay although it might take a few decades in ×4.
— Yoha (talk) 14:12, 26 December 2014 (CST)
- But that's the issue I have (and I've written it down here to link anyone to it if needed): The atmospheric height stops at 1/100000th of the sea level pressure and not at 1/100000th atm, which would make more sense because the drag depends at on the pressure and not on the “relative sea level pressure”. I know that at about 70 km the atmosphere suddenly stops. Directly at the border the pressure is 10⁻⁶ atm so you'd expect the same pressure at the border around every other body.
- You also stated that it makes calculations easier but I'm not sure about that. I obviously don't know how Squad does calculate it, but making it an absolute shouldn't be as hard as it just requires an additional division (instead of 10⁻⁶ it's 10⁻⁶/p_0). — xZise [talk] 15:54, 26 December 2014 (CST)
- I agree that stopping the atmosphere at 10⁻⁶ atm everywhere would make more sense. As for the simpler computations, I was referring to orbits above the threshold, for which we have closed formula for pretty much everything as long as we ignore drag. Indeed, having a relative threshold instead of an absolute one for the top of the atmosphere does not simplify anything at all. So you can consider that KSP is at fault there, even if that is not really “wrong”, just a poor choice. — Yoha (talk) 03:48, 27 December 2014 (CST)
- Well I think it's a bug because it looks like someone just forgot the division. And of course it can't be ”wrong”, because maybe physics in KSP work differently which would explain the different “border”-pressures. It is “just” an inconsistency which doesn't make sense. — xZise [talk] 13:09, 27 December 2014 (CST)
- I agree that stopping the atmosphere at 10⁻⁶ atm everywhere would make more sense. As for the simpler computations, I was referring to orbits above the threshold, for which we have closed formula for pretty much everything as long as we ignore drag. Indeed, having a relative threshold instead of an absolute one for the top of the atmosphere does not simplify anything at all. So you can consider that KSP is at fault there, even if that is not really “wrong”, just a poor choice. — Yoha (talk) 03:48, 27 December 2014 (CST)
“you know could construct an example with an sea level pressure of 10⁺⁶ atm resulting in a pressure of 1 atm at the “atmospheric height” which is the pressure at Kerbin's sea level.”
Absolutely, since the threshold is relative the pressure at sea level, a planet could be conceived where the pressure discontinuity would be huge. Using an absolute threshold would make more sense. — Yoha (talk) 14:12, 26 December 2014 (CST) The problem is not about the magnitude of the discontinuity (although it would be 100000 times larger) but that the inconsistency in the game. There are several simplifications (e.g. no n-body or no drag when not active) but those are understandable because they are getting complex fast or are complex as they are. And I don't think this is currently a simplification but just overlooked or something. The fix should be also relatively easy and could be done with any save-game breaking change (because in theory craft could be to low with the new calculations). — xZise [talk] 15:54, 26 December 2014 (CST)