https://wiki.kerbalspaceprogram.com/api.php?action=feedcontributions&user=Tavert&feedformat=atomKerbal Space Program Wiki - User contributions [en]2024-03-28T14:07:33ZUser contributionsMediaWiki 1.29.0https://wiki.kerbalspaceprogram.com/index.php?title=Talk:Eve&diff=26115Talk:Eve2013-09-20T03:49:47Z<p>Tavert: </p>
<hr />
<div>== liquid ==<br />
the article says that the oceans/lakes are unlikely to be water. i'm not sure if outdated data is involved here, but wolfram alpha says that [https://www.wolframalpha.com/input/?i=boiling+point+of+water+507+kpa the boiling point of water at 507 kPa] is 2.5° ''higher'' than the max surface temperature (150°) listed in the infobox, suggesting that water could very well exist as a liquid on eve. i'll amend that part unless there are any objections. --[[User:Ant|ant]] ([[User talk:Ant|☎]]) 15:45, 20 August 2013 (CDT)<br />
:Your Wolfram Alpha link doesn't seem to work, but assuming that you did it right I guess water wouldn't boil. Also maybe the temperature values are invalid or outdated. And your link to the talk page was incorrect. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 17:19, 20 August 2013 (CDT)<br />
::weird... [http://www.wolframalpha.com/input/?i=water+at+507+kpa this] should work. and thanks for the heads-up, something must have gone screwy when i ticked the "treat as wikitext" box. --[[User:Ant|ant]] ([[User talk:Ant|☎]]) 18:23, 20 August 2013 (CDT)<br />
:::Okay this link is now working. And if I read correctly salty water has an even higher boiling point (unfortunately I can't get wolfram alpha to calculate that value). I'll send probe(s) to Eve and check the actual pressure and temperature levels, and if the values in the infobox are correct the section should be rewritten. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 12:37, 21 August 2013 (CDT)<br />
== Infiniglider glitch ==<br />
A few days ago, I flew around Eve by exploiting the game mechanics so that I could use control surfaces for propulsion and stay aloft indefinitely (on a similar principle to the Duna probes seen in [http://www.youtube.com/watch?v=H0YhNvtcXLU this video]. Should this be mentioned as a legitimate strategy, or should it stay off the page as a glitch exploit? [[User:Ammonia ocean|Ammonia ocean]] ([[User talk:Ammonia ocean|talk]]) 22:14, 3 September 2013 (CDT)<br />
:I would prefer an article like [[infinite glider]] and describe the “physics” there. Only note that those glider work in Eve's atmosphere, because this is not a special property of Eve. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:24, 4 September 2013 (CDT)<br />
:Okay, I made some additions to the [[infinite glider]] article. Let me know what else you need added. [[User:Ammonia ocean|Ammonia ocean]] ([[User talk:Ammonia ocean|talk]]) 22:09, 4 September 2013 (CDT)<br />
::Thanks for your additions, it now looks nice. I moved your image to the left like [[plane]], [[rocket]] or [[spaceplane]] do it too. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:02, 5 September 2013 (CDT)<br />
::No problem. Thanks for moving the image, I was wondering how to do that. [[User:Ammonia ocean|Ammonia ocean]] ([[User talk:Ammonia ocean|talk]]) 09:52, 5 September 2013 (CDT)<br />
<br />
== Delta-V Required table ==<br />
The recently added table requires citation, and it should be emphasized that any such numbers are intended as minimum values, as ascent delta-V is both craft (particularly TWR) and trajectory dependent. These should be obvious facts, but if they are left unstated folks will be asking why their rockets didn't make orbit despite having as much dV as the wiki said they needed. Best to address that question along with the data.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 22:49, 19 September 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=J-33_%22Wheesley%22_Turbofan_Engine&diff=25299J-33 "Wheesley" Turbofan Engine2013-09-03T04:51:53Z<p>Tavert: Jet engine thrust depends on speed, not altitude, see velocityCurve in part.cfg's. Isp depends on altitude.</p>
<hr />
<div>{{:Basic Jet Engine/Box}}<br />
<br />
The '''Basic Jet Engine''' is an [[Jet engine|air-breathing engine]] which uses [[liquid fuel]] and [[intake air]].<br />
<br />
== Usage ==<br />
This engine cannot function effectively at a height of much more than 5&nbsp;km, but it is much more efficient to use jet engines in the lower atmosphere. The basic jet engine works at full power while stationary, whereas the [[TurboJet Engine]] won't get up to full power until it reaches 1000 meters/second surface velocity. This makes it ideal for low-flying [[plane|aircraft]], as the engine has a thrust of about 100 kilonewtons on the runway and an I<sub>sp</sub> of about 850. However, for high-flying aircraft, the TurboJet engine should still get you off the runway.<br />
<br />
Note that currently, the only planets which these engines will work on is [[Kerbin]], and Jool's moon [[Laythe]].<br />
<br />
== Description ==<br />
{{Quote<br />
|A highly advanced replacement model of the Ax-300 series of engines. This new model features the latest in thrust vectoring and compression technology. This engine is suitable for normal crusing speeds and altitudes.<br />
|C7 Aerospace Division}}<br />
<br />
Created by: C. Jenkins<br />
<br />
== Changes ==<br />
;{{Version|0.15|}}<br />
* Initial Release <br />
<br />
{{Parts}}<br />
[[Category:Jet engines]]</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Atmosphere&diff=24961Atmosphere2013-08-28T22:12:47Z<p>Tavert: revise poor English</p>
<hr />
<div>{| class="wikitable float-right"<br />
! colspan="2" | Planets<br />
|-<br />
| style="border-right:0px;" | [[File:TinyEve.png|16px]]<br />
| style="border-left:0px;" | [[Eve]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyKerbin.png|16px]]<br />
| style="border-left:0px;" | [[Kerbin]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyDuna.png|16px]]<br />
| style="border-left:0px;" | [[Duna]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyJool.png|16px]]<br />
| style="border-left:0px;" | [[Jool]]<br />
|-<br />
! colspan="2" | Moons<br />
|-<br />
| style="border-right:0px;" | [[File:TinyLaythe.png|16px]]<br />
| style="border-left:0px;" | [[Laythe]]<br />
|}<br />
<br />
The '''atmosphere''' of a celestial body slows the movement of any object passing through it, a force known as atmospheric drag (or simply '''drag'''). An atmosphere also allows for aerodynamic lift. The celestial bodies with atmospheres are the planets [[Eve]], [[Kerbin]], [[Duna]] and [[Jool]], as well as [[Laythe]], a moon of Jool. Only [[Kerbin]] and [[Laythe]] have atmospheres that contain oxygen.<br />
<br />
Atmospheric pressure diminishes exponentially with increasing altitude. An atmosphere's ''scale height'' is the distance over which atmospheric pressure changes as a factor of ''e'', or 2.718. For example, Kerbin's atmosphere has a scale height of 5000 m, meaning the atmospheric pressure at altitude ''n'' is 2.718 times greater than the pressure at altitude ''n'' + 5000.<br />
<br />
Atmospheres vary in temperature, though this has no bearing on gameplay.<br />
<br />
Atmospheres allow [[aerobraking]] and easier landing. When an atmosphere contains oxygen, it allows [[jet engine]]s to work. However, an atmosphere makes taking off from a planet more difficult and increases the minimum stable orbit altitude.<br />
<br />
== Drag ==<br />
[[File:Ml16-XL_parachute.JPG|thumb|right|A Mk1-2 pod with a Mk16-XL parachute being slowed by drag in Kerbin's atmosphere.]]<br />
<br />
In the game, the force of atmospheric drag (''F<sub>D</sub>'') is modeled as follows:<ref>http://forum.kerbalspaceprogram.com/showthread.php/5235-Atmospheric-drag?p=88804&viewfull=1#post88804</ref><br />
<br />
: <math>F_D = 0.5\, \rho\, v^2\, d\, A</math><br />
<br />
where ''&rho;'' is the atmospheric density (kg/m<sup>3</sup>), ''v'' is the ship's velocity (m/s), ''d'' is the coefficient of drag (dimensionless), and ''A'' is the [[w:cross section (geometry)|cross-sectional area]] (m<sup>2</sup>).<br />
<br />
Note that the cross-sectional area is not actually calculated in the game. It is instead assumed that it is directly proportional to the mass, which is an unrealistic simplification made by KSP. The parameter [[API:FlightGlobals|FlightGlobals]].DragMultiplier indicates that the proportionality ratio is 0.008 m<sup>2</sup>/kg, so:<br />
<br />
: <math>A = 0.008 \cdot m</math><br />
<br />
where ''m'' is the ship's mass (kg).<br />
<br />
The atmospheric density ''&rho;'' is directly proportional to atmospheric pressure (''p'' of unit ''atm''), which is a function of altitude, the atmosphere's pressure at altitude 0 (''p<sub>0</sub>''), and scale height (''H''):<br />
<br />
: <math>p = p_0 \cdot e^\frac{-altitude}{H}</math><br />
<br />
: <math>\rho = 1.2230948554874 \cdot p</math><br />
<br />
where p here is in units atm, and ''&rho;'' in kg/m<sup>3</sup>. The 1.2230948554874 kg/m<sup>3</sup> value for density at 1 atmosphere (sea level on Kerbin) is given by [[API:FlightGlobals|FlightGlobals]].getAtmDensity(1.0).<br />
<br />
The coefficient of drag (''d'') is calculated as the mass-weighted average of the max_drag values of all [[parts]] on the ship. For most ships without deployed parachutes, ''d'' will be very near 0.2, since this is the max_drag value of the vast majority of parts.<br />
<br />
As an example, the coefficient of drag for a craft consisting simply of a [[Mk1-2 Command Pod]] (mass 4, drag 0.2) and a deployed [[Mk16-XL Parachute]] (mass 0.3, drag 500) is:<br />
<br />
: <math>\frac{4 \cdot 0.2 + 0.3 \cdot 500}{4 + 0.3} = 35.07</math><br />
<br />
== Terminal velocity ==<br />
The [[w:terminal velocity|terminal velocity]] of an object falling through an atmosphere is the velocity at which the force of gravity is equal to the force of drag. Terminal velocity changes as a function of altitude. Given enough time, an object falling into the atmosphere will slow to terminal velocity and then remain at terminal velocity for the rest of its fall.<br />
<br />
Terminal velocity is important because:<br />
# It describes the amount of velocity which a spacecraft must burn away when it is close to the ground.<br />
# It represents the speed at which a ship should be traveling upward during a fuel-optimal ascent.<br />
<br />
The force of gravity (''F<sub>G</sub>'') is:<br />
<br />
: <math>F_G = m \cdot a = m \cdot \frac{GM}{r^2}</math><br />
<br />
where ''m'' is still the ship's mass, ''G'' is the [[Template:G|gravitational constant]], ''M'' is the mass of the planet, and ''r'' is the distance from the ''center'' of the planet to the falling object.<br />
<br />
To find terminal velocity, we set ''F<sub>G</sub>'' equal to ''F<sub>D</sub>'':<br />
<br />
: <math>\begin{align}<br />
m \cdot \frac{GM}{r^2} &= 0.5 \cdot \rho \cdot v^2 \cdot d \cdot (0.008 \cdot m) \\<br />
\frac{GM}{r^2} &= 0.004 \cdot \rho \cdot v^2 \cdot d \\<br />
v &= v_T = \sqrt{\frac{250 \cdot GM}{r^2 \cdot \rho \cdot d}}<br />
\end{align}</math><br />
<br />
Assuming ''d'' is 0.2 (which is a good approximation, provided parachutes are not in use), this simplifies to:<br />
<br />
: <math>v_T = \sqrt{\frac{1250\, GM}{r^2\, \rho}}</math><br />
<br />
For the Mk16 pod and parachute example pictured above, the drag coefficient is 35.07, so its terminal velocity at sea level on Kerbin (which is 600 km from Kerbin's center) is:<br />
<br />
: <math>v_T = \sqrt{\frac{250\, GM}{r^2\, \rho \cdot 35.07}}</math><br />
<br />
: <math>\rho = 1.2230948554874 \cdot 1 \cdot e^\frac{-0}{5000}</math><br />
<br />
: <math>v_T = \sqrt{\frac{250 \cdot 6.674 \cdot 10^{-11} \cdot 5.2915793 \cdot 10^{22}}{600000^2 \cdot 1.2230948554874 \cdot 35.07}} = 7.56 \operatorname{m/s}</math><br />
<br />
=== Examples ===<br />
{| class="wikitable"<br />
!rowspan=2 valign=bottom| Altitude (m) ||colspan=6| v<sub>T</sub> (m/s)<br />
|-<br />
! Eve !! Kerbin !! Duna !! Jool !! Laythe<br />
|-<br />
| 0 || {{sigfigs|{{VT | planet=Eve | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 0}}|5}}<br />
|-<br />
| 100 || {{sigfigs|{{VT | planet=Eve | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 100}}|5}}<br />
|-<br />
| 1000 || {{sigfigs|{{VT | planet=Eve | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 1000}}|5}}<br />
|-<br />
| 10000 || {{sigfigs|{{VT | planet=Eve | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Duna | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Jool | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt=10000}}|5}}<br />
|}<br />
<br />
== On-rails physics ==<br />
If a ship is "on rails" (meaning it's further than 2.25&nbsp;km from the actively-controlled ship) and its orbit passes through a planet's atmosphere, one of two things will happen based on atmospheric pressure at the ship's altitude:<br />
<br />
* below 0.01&nbsp;atm: no atmospheric drag will occur &mdash; the ship will be completely unaffected<br />
* 0.01&nbsp;atm or above: the ship will disappear<br />
<br />
The following table gives the altitude of this 0.01&nbsp;atm threshold for each celestial body with an atmosphere:<br />
{| class="wikitable"<br />
|-<br />
! Body || Altitude (m)<br />
|-<br />
| [[Eve]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Eve}}}}<br />
|-<br />
| [[Kerbin]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Kerbin}}}}<br />
|-<br />
| [[Duna]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Duna}}}}<br />
|-<br />
| [[Jool]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Jool}}}}<br />
|-<br />
| [[Laythe]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Laythe}}}}<br />
|}<br />
<br />
== Atmospheric height ==<br />
The atmospheric height depend on the scale height of the celestial body and is where 0.000001<sup>th</sup> (0.0001&nbsp;%) of the surface pressure is remaining so the atmospheric pressure at the border isn't constant. Technically a craft in Jool's orbit can get lower into the atmosphere (or the atmosphere starts from a higher pressure).<br />
:<math>alt_{\text{atmospheric height}} = -ln\left(10^{-6}\right) \cdot \text{scale height}</math><br />
:<math>p_{\text{atmospheric height}} = p_0 \cdot 10^{-6}</math><br />
Kerbin's atmosphere ends at 0.000001&nbsp;atm and to calculate where the other celestial bodies should have the atmospheric height:<br />
:<math>alt_{\text{atmospheric height (real)}} = -ln\left(\frac{10^{-6}}{p_0}\right) \cdot \text{scale height}</math><br />
<br />
== Notes ==<br />
<references /></div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Atmosphere&diff=24794Atmosphere2013-08-25T15:50:07Z<p>Tavert: whoops, missed a confusing "A = 1" comment</p>
<hr />
<div>{| class="wikitable float-right"<br />
! colspan="2" | Planets<br />
|-<br />
| style="border-right:0px;" | [[File:TinyEve.png|16px]]<br />
| style="border-left:0px;" | [[Eve]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyKerbin.png|16px]]<br />
| style="border-left:0px;" | [[Kerbin]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyDuna.png|16px]]<br />
| style="border-left:0px;" | [[Duna]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyJool.png|16px]]<br />
| style="border-left:0px;" | [[Jool]]<br />
|-<br />
! colspan="2" | Moons<br />
|-<br />
| style="border-right:0px;" | [[File:TinyLaythe.png|16px]]<br />
| style="border-left:0px;" | [[Laythe]]<br />
|}<br />
<br />
The '''atmosphere''' of a celestial body slows the movement of any object passing through it, a force known as atmospheric drag (or simply '''drag'''). An atmosphere also allows for aerodynamic lift. The celestial bodies with atmospheres are the planets [[Eve]], [[Kerbin]], [[Duna]] and [[Jool]], as well as [[Laythe]], a moon of Jool. Only [[Kerbin]] and [[Laythe]] have atmospheres that contain oxygen.<br />
<br />
Atmospheric pressure diminishes exponentially with increasing altitude. An atmosphere's ''scale height'' is the distance over which atmospheric pressure changes as a factor of ''e'', or 2.718. For example, Kerbin's atmosphere has a scale height of 5000 m, meaning the atmospheric pressure at altitude ''n'' is 2.718 times greater than the pressure at altitude ''n'' + 5000.<br />
<br />
Atmospheres vary in temperature, though this has no bearing on gameplay.<br />
<br />
Atmospheres allow [[aerobraking]] and easier landing. When an atmosphere contains oxygen, it allows [[jet engine]]s to work. However, an atmosphere makes taking off from a planet more difficult and increases the minimum stable orbit altitude.<br />
<br />
== Drag ==<br />
[[File:Ml16-XL_parachute.JPG|thumb|right|A Mk1-2 pod with a Mk16-XL parachute being slowed by drag in Kerbin's atmosphere.]]<br />
<br />
In the game, the force of atmospheric drag (''F<sub>D</sub>'') is modeled as follows:<ref>http://forum.kerbalspaceprogram.com/showthread.php/5235-Atmospheric-drag?p=88804&viewfull=1#post88804</ref><br />
<br />
: <math>F_D = 0.5\, \rho\, v^2\, d\, A</math><br />
<br />
where ''&rho;'' is the atmospheric density (kg/m<sup>3</sup>), ''v'' is the ship's velocity (m/s), ''d'' is the coefficient of drag (dimensionless), and ''A'' is the [[w:cross section (geometry)|cross-sectional area]] (m<sup>2</sup>).<br />
<br />
Note that the cross-sectional area is not actually calculated in the game. It is instead assumed that it is directly proportional to the mass. The parameter [[API:FlightGlobals|FlightGlobals]].DragMultiplier indicates that the proportionality ratio is 0.008 m<sup>2</sup>/kg, so:<br />
<br />
: <math>A = 0.008 \cdot m</math><br />
<br />
where ''m'' is the ship's mass (kg).<br />
<br />
The atmospheric density ''&rho;'' is directly proportional to atmospheric pressure (''p'' of unit ''atm''), which is a function of altitude, the atmosphere's pressure at altitude 0 (''p<sub>0</sub>''), and scale height (''H''):<br />
<br />
: <math>p = p_0 \cdot e^\frac{-altitude}{H}</math><br />
<br />
: <math>\rho = 1.2230948554874 \cdot p</math><br />
<br />
where p here is in units atm, and ''&rho;'' in kg/m<sup>3</sup>. The 1.2230948554874 kg/m<sup>3</sup> value for density at 1 atmosphere (sea level on Kerbin) is given by [[API:FlightGlobals|FlightGlobals]].getAtmDensity(1.0).<br />
<br />
The coefficient of drag (''d'') is calculated as the mass-weighted average of the max_drag values of all [[parts]] on the ship. For most ships without deployed parachutes, ''d'' will be very near 0.2, since this is the max_drag value of the vast majority of parts.<br />
<br />
As an example, the coefficient of drag for a craft consisting simply of a [[Mk1-2 Command Pod]] (mass 4, drag 0.2) and a deployed [[Mk16-XL Parachute]] (mass 0.3, drag 500) is:<br />
<br />
: <math>\frac{4 \cdot 0.2 + 0.3 \cdot 500}{4 + 0.3} = 35.07</math><br />
<br />
== Terminal velocity ==<br />
The [[w:terminal velocity|terminal velocity]] of an object falling through an atmosphere is the velocity at which the force of gravity is equal to the force of drag. Terminal velocity changes as a function of altitude. Given enough time, an object falling into the atmosphere will slow to terminal velocity and then remain at terminal velocity for the rest of its fall.<br />
<br />
Terminal velocity is important because:<br />
# It describes the amount of velocity which a spacecraft must burn away when it is close to the ground.<br />
# It represents the speed at which a ship should be traveling upward during a fuel-optimal ascent.<br />
<br />
The force of gravity (''F<sub>G</sub>'') is:<br />
<br />
: <math>F_G = m\, a = m\, \frac{GM}{r^2}</math><br />
<br />
where ''m'' is still the ship's mass, ''G'' is the [[Template:G|gravitational constant]], ''M'' is the mass of the planet, and ''r'' is the distance from the ''center'' of the planet to the falling object.<br />
<br />
To find terminal velocity, we set ''F<sub>G</sub>'' equal to ''F<sub>D</sub>'':<br />
<br />
: <math>m\, \frac{GM}{r^2} = 0.5\, \rho\, v^2\, d\, (0.008 \cdot m)</math><br />
<br />
: <math>\frac{GM}{r^2} = 0.004\, \rho\, v^2\, d</math><br />
<br />
: <math>v = v_T = \sqrt{\frac{250\, GM}{r^2\, \rho\, d}}</math><br />
<br />
Assuming ''d'' is 0.2 (which is a good approximation, provided parachutes are not in use), this simplifies to:<br />
<br />
: <math>v_T = \sqrt{\frac{1250\, GM}{r^2\, \rho}}</math><br />
<br />
For the Mk16 pod and parachute example pictured above, the drag coefficient is 35.07, so its terminal velocity at sea level on Kerbin (which is 600 km from Kerbin's center) is:<br />
<br />
: <math>v_T = \sqrt{\frac{250\, GM}{r^2\, \rho \cdot 35.07}}</math><br />
<br />
: <math>\rho = 1.2230948554874 \cdot 1 \cdot e^\frac{-0}{5000}</math><br />
<br />
: <math>v_T = \sqrt{\frac{250 \cdot 6.674 \cdot 10^{-11} \cdot 5.2915793 \cdot 10^{22}}{600000^2 \cdot 1.2230948554874 \cdot 35.07}} = 7.56 \operatorname{m/s}</math><br />
<br />
=== Examples ===<br />
{| class="wikitable"<br />
!rowspan=2 valign=bottom| Altitude (m) ||colspan=6| v<sub>T</sub> (m/s)<br />
|-<br />
! Eve !! Kerbin !! Duna !! Jool !! Laythe<br />
|-<br />
| 0 || {{sigfigs|{{VT | planet=Eve | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 0}}|5}}<br />
|-<br />
| 100 || {{sigfigs|{{VT | planet=Eve | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 100}}|5}}<br />
|-<br />
| 1000 || {{sigfigs|{{VT | planet=Eve | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 1000}}|5}}<br />
|-<br />
| 10000 || {{sigfigs|{{VT | planet=Eve | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Duna | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Jool | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt=10000}}|5}}<br />
|}<br />
<br />
== On-rails physics ==<br />
If a ship is "on rails" (meaning it's further than 2.25&nbsp;km from the actively-controlled ship) and its orbit passes through a planet's atmosphere, one of two things will happen based on atmospheric pressure at the ship's altitude:<br />
<br />
* below 0.01&nbsp;atm: no atmospheric drag will occur &mdash; the ship will be completely unaffected<br />
* 0.01&nbsp;atm or above: the ship will disappear<br />
<br />
The following table gives the altitude of this 0.01&nbsp;atm threshold for each celestial body with an atmosphere:<br />
{| class="wikitable"<br />
|-<br />
! Body || Altitude (m)<br />
|-<br />
| [[Eve]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Eve}}}}<br />
|-<br />
| [[Kerbin]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Kerbin}}}}<br />
|-<br />
| [[Duna]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Duna}}}}<br />
|-<br />
| [[Jool]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Jool}}}}<br />
|-<br />
| [[Laythe]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Laythe}}}}<br />
|}<br />
<br />
== Atmospheric height ==<br />
The atmospheric height depend on the scale height of the celestial body and is where 0.000001<sup>th</sup> (0.0001&nbsp;%) of the surface pressure is remaining so the atmospheric pressure at the border isn't constant. Technically a craft in Jool's orbit can get lower into the atmosphere (or the atmosphere starts from a higher pressure).<br />
:<math>alt_{\text{atmospheric height}} = -ln\left(10^{-6}\right) \cdot \text{scale height}</math><br />
:<math>p_{\text{atmospheric height}} = p_0 \cdot 10^{-6}</math><br />
Kerbin's atmosphere ends at 0.000001&nbsp;atm and to calculate where the other celestial bodies should have the atmospheric height:<br />
:<math>alt_{\text{atmospheric height (real)}} = -ln\left(\frac{10^{-6}}{p_0}\right) \cdot \text{scale height}</math><br />
<br />
== Notes ==<br />
<references /></div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Jet_engine&diff=24775Jet engine2013-08-25T09:27:27Z<p>Tavert: Jet engine thrust depends on speed, not altitude, see velocityCurve in part.cfg's. Isp depends on altitude.</p>
<hr />
<div>[[Image:JetEngine.png|right|thumb|A jet engine from KSP 0.18]]<br />
<br />
A '''jet engine''' is an air-breathing engine which uses onboard fuel and combusts it with the oxygen drawn from the atmosphere. Due to the lower air pressure at higher altitudes its thrust output varies accordingly.<br />
<br />
== Usage ==<br />
Unlike rocket engines, a jet draws oxygen from the atmosphere rather than taking it from an on-board tank. This is represented in-game by a much lower rate of fuel consumption. Depending on the aircraft, jet engines can produce thrust at altitudes greater than 20km. Jet engines do not perform in the same manner as rocket engines. Since Jet engines rely on a turbine to compress fuel and atmosphere together in order to produce thrust, it takes time to change speed. Also if the throttle or air flow are too low the engine will stall. By contrast, rocket engines give instant response to control input.<br />
<br />
As of [[Version_History#v0.18.0|version 0.18]] jet engines require air to run, provided by [[air intake]]s.<br />
<br />
== Advantages ==<br />
<br />
* Provides excellent fuel efficiency within an atmosphere<br />
* All current jet engines provide thrust vectoring for greater maneuverability<br />
* Excellent power to weight ratio<br />
<br />
== Disadvantages ==<br />
<br />
* Cannot be used outside of an atmosphere that contains oxygen. In current version, it means they only function on [[Kerbin]] and [[Laythe]].<br />
* Thrust output changes depending on speed<br />
* Engine requires time to spool up to maximum thrust potential<br />
<br />
== Available jet engines ==<br />
{{Stats Table Jet Engines}}<br />
<br />
== See also ==<br />
* {{Wikipedia}}<br />
<br />
[[Category:Parts]]<br />
[[Category:Jet engines| ]]</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24771Talk:Atmosphere2013-08-25T09:13:06Z<p>Tavert: </p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)<br />
:Maybe leave a note on [[User talk:Richardo 5|Richardo 5]]'s talk page. (S)He made “measurements” and then changed the constants. Di you measure a constant of 1.2002 at the height of the launch pad? Because I get a different value: e<sup>-68m/5000m</sup>×1.2230948554874=1.20657337<br />
:When I'm not mistaken, this would be the constant when you have 1 atm at the launch pad. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:06, 24 August 2013 (CDT)<br />
::It looks like it was [[User talk:Nulvinge|Nulvinge]] who added the ideal gas law equation but apparently without checking the numbers, then [[User talk:Richardo 5|Richardo 5]] plugged Nulvinge's ideal gas law numbers into a calculator and that's where the 1.2002 number came from. Here's a quick screenshot with MechJeb's density readout showing on the launch pad at 1.206 (close to your prediction), but it's kind of a short rocket so maybe on tall launch clamps you could get to 1.2002? http://i.imgur.com/FKz6PaY.jpg And part of my point was that you don't have 1 atm pressure at the launch pad, you have roughly 0.986 atm.<br />
::I think it should be noted rather than "the way the Kerbal universe works" that the 0.008 number comes from the API function FlightGlobals.DragMultiplier, and likewise with sea-level density from FlightGlobals.getAtmDensity(1.0), and these API functions are what mods like MechJeb and Kerbal Engineer use in their source to measure (and predict in MechJeb's case) drag and density. If these numbers from the API functions didn't reflect the game physics, then the atmospheric-landing and aerobraking predictions in MechJeb would have noticeably larger errors. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:39, 24 August 2013 (CDT)<br />
:::Ah nice. Yeah I guess this could be changed to the official numbers, maybe a little text that the density on the launch pad might differ. Sitting on the launch pad at about 68 m (AMSL) the measurement point should be about 26 meters above the launch pad (or 94 m AMSL). — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:23, 24 August 2013 (CDT)<br />
::::Having looked through the history and tracked it down to one user adding the ideal gas law without checking the numbers and another user replacing the numbers assuming the ideal gas law equation was reflecting KSP physics, I don't think the 1.2002 number, or discussion about the launch pad altitude, or the ideal gas law are necessary at all.<br />
::::I think it makes the most sense to say A = 0.008 (m^2/kg) * mass, and note that the 0.008 conversion factor comes from FlightGlobals.DragMultiplier. The density varies proportionally to pressure (with FlightGlobals.getAtmDensity(1.0) = 1.2230948554874), which is consistent with an isothermal atmosphere assumption so the temperature sensor readout is not at all related to pressure or density or drag, and the ideal gas law does not really hold for atmospheres in KSP.<br />
::::I could try going through and revising, but I'm not too familiar with wiki formatting or etiquette so I wanted to discuss first. It's a bit of a long article with numbers showing up in several places, any lingering inconsistencies from missed numbers would confuse people.<br />
::::A separate quibble that applies here and a few other places on the wiki, 1 atmosphere is defined by the CGPM as exactly 101325 Pascals. 101 kPa is a rounding, 101.327 kPa (which I see in a few places) is closer but also wrong. It's possible that Kerbal atmospheres means a different thing, but I think the original source of the 101.327 number is from this spreadsheet: https://docs.google.com/spreadsheet/ccc?key=0AuySrGPsDeq2dFdaS19xc2lobGc2aWNXUkJsZlVtWFE#gid=1 However I can't seem to find any mention of that number in the source forum thread or any of the linked pastebin data. FlightGlobals.getStaticPressure returns in atmospheres, and I guess the 0-altitude pressures for the bodies other than Kerbin were found from other sources (luckily they're all round numbers). Thanks for responding to this. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 06:22, 24 August 2013 (CDT)<br />
:::::I also noticed that the {{Tl|Infobox/Body}} uses 101.327 kPa/atm but nobody [[Template_talk:Infobox/Body#Difference_between_.E2.80.9Creal.E2.80.9D_atm_and_KSP_atm_pressure|replied on the talk page]].<br />
:::::Unfortunately I don't know that much about pressures and such, so I don't know how the ideal gas law is applying here (or not). But I guess the main problem is [http://wiki.kerbalspaceprogram.com/w/index.php?title=Atmosphere&diff=22309&oldid=20448 this edit], now [https://www.google.de/search?q=101.325+kPa/%281+atm*287.058+J/%28kg+K%29*293.15K%29 Google returns] (even with the correct value kPa for 1 atm) a totally different value, so I don't know where the error is: In the formula, the constants or even using the formula. Interestingly if you multiply that constant with 1 atm you get this 1.2002 (when using 101 kPa, or 1.2041 when using 101.325 kPa).<br />
:::::But nevertheless I would recommend to fix all occurrences of the incorrect constant. And the formula should also be removed as long as we don't know what is wrong with it. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 08:13, 24 August 2013 (CDT)<br />
:::::Oh and yes, those 0.008 should be removed from the pressure/density conversion formula, as MechJeb shows. I guess you want to add this coefficient to the first formula calculating the force? — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 08:24, 24 August 2013 (CDT)<br />
::::::I added the 0.008 factor as if it reflects the proportionality ratio between mass and cross-sectional area. I'm not sure if this is exactly the intent of the developers, but it makes sense to me.--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:13, 25 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Atmosphere&diff=24769Atmosphere2013-08-25T09:10:54Z<p>Tavert: Fix incorrect sea-level density, move 0.008 drag multiplier from density to area calculation, remove ideal gas law which doesn't apply to Kerbin's atmosphere. See talk page.</p>
<hr />
<div>{| class="wikitable" style="float:right;margin:0.5em;"<br />
! colspan="2" | Planets<br />
|-<br />
| style="border-right:0px;" | [[File:TinyEve.png|16px]]<br />
| style="border-left:0px;" | [[Eve]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyKerbin.png|16px]]<br />
| style="border-left:0px;" | [[Kerbin]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyDuna.png|16px]]<br />
| style="border-left:0px;" | [[Duna]]<br />
|-<br />
| style="border-right:0px;" | [[File:TinyJool.png|16px]]<br />
| style="border-left:0px;" | [[Jool]]<br />
|-<br />
! colspan="2" | Moons<br />
|-<br />
| style="border-right:0px;" | [[File:TinyLaythe.png|16px]]<br />
| style="border-left:0px;" | [[Laythe]]<br />
|}<br />
<br />
The '''atmosphere''' of a celestial body slows the movement of any object passing through it, a force known as atmospheric drag (or simply '''drag'''). An atmosphere also allows for aerodynamic lift. The celestial bodies with atmospheres are the planets [[Eve]], [[Kerbin]], [[Duna]] and [[Jool]], as well as [[Laythe]], a moon of Jool. Only [[Kerbin]] and [[Laythe]] have atmospheres that contain oxygen.<br />
<br />
Atmospheric pressure diminishes exponentially with increasing altitude. An atmosphere's ''scale height'' is the distance over which atmospheric pressure changes as a factor of ''e'', or 2.718. For example, Kerbin's atmosphere has a scale height of 5000 m, meaning the atmospheric pressure at altitude ''n'' is 2.718 times <br />
greater than the pressure at altitude ''n'' + 5000.<br />
<br />
Atmospheres vary in temperature, though this has no bearing on gameplay.<br />
<br />
Atmospheres allow [[aerobraking]] and easier landing. When an atmosphere contains oxygen, it allows jet engines to work. However, an atmosphere makes taking off from a planet more difficult and increases the minimum stable orbit altitude.<br />
<br />
== Drag ==<br />
[[File:Ml16-XL_parachute.JPG|thumb|right|A Mk1-2 pod with a Mk16-XL parachute being slowed by drag in Kerbin's atmosphere.]]<br />
<br />
In the game, the force of atmospheric drag (''F<sub>D</sub>'') is modeled as follows:<ref>http://forum.kerbalspaceprogram.com/showthread.php/5235-Atmospheric-drag?p=88804&viewfull=1#post88804</ref><br />
<br />
: <math>F_D = 0.5\, \rho\, v^2\, d\, A</math><br />
<br />
where ''&rho;'' is the atmospheric density (kg/m<sup>3</sup>), ''v'' is the ship's velocity (m/s), ''d'' is the coefficient of drag (dimensionless), and ''A'' is the [[w:cross section (geometry)|cross-sectional area]] (m<sup>2</sup>).<br />
<br />
Note that the cross-sectional area is not actually calculated in the game. It is instead assumed that it is directly proportional to the mass. The parameter [[API:FlightGlobals|FlightGlobals]].DragMultiplier indicates that the proportionality ratio is 0.008 m<sup>2</sup>/kg, so:<br />
<br />
: <math>A = 0.008 \cdot m</math><br />
<br />
where ''m'' is the ship's mass (kg).<br />
<br />
The atmospheric density ''&rho;'' is directly proportional to atmospheric pressure (''p'' of unit ''atm''), which is a function of altitude, the atmosphere's pressure at altitude 0 (''p<sub>0</sub>''), and scale height (''H''):<br />
<br />
: <math>p = p_0 \cdot e^{-altitude / H}</math><br />
<br />
: <math>\rho = 1.2230948554874 \cdot p</math><br />
<br />
where p here is in units atm, and ''&rho;'' in kg/m<sup>3</sup>. The 1.2230948554874 kg/m<sup>3</sup> value for density at 1 atmosphere (sea level on Kerbin) is given by [[API:FlightGlobals|FlightGlobals]].getAtmDensity(1.0).<br />
<br />
The coefficient of drag (''d'') is calculated as the mass-weighted average of the max_drag values of all [[parts]] on the ship. For most ships without deployed parachutes, ''d'' will be very near 0.2, since this is the max_drag value of the vast majority of parts.<br />
<br />
As an example, the coefficient of drag for a craft consisting simply of a [[Mk1-2 Command Pod]] (mass 4, drag 0.2) and a deployed [[Mk16-XL Parachute]] (mass 0.3, drag 500) is:<br />
<br />
: <math>\frac{4 \cdot 0.2 + 0.3 \cdot 500}{4 + 0.3} = 35.07</math><br />
<br />
== Terminal velocity ==<br />
The [[w:terminal velocity|terminal velocity]] of an object falling through an atmosphere is the velocity at which the force of gravity is equal to the force of drag. Terminal velocity changes as a function of altitude. Given enough time, an object falling into the atmosphere will slow to terminal velocity and then remain at terminal velocity for the rest of its fall.<br />
<br />
Terminal velocity is important because:<br />
# It describes the amount of velocity which a spacecraft must burn away when it is close to the ground.<br />
# It represents the speed at which a ship should be traveling upward during a fuel-optimal ascent.<br />
<br />
The force of gravity (''F<sub>G</sub>'') is:<br />
<br />
: <math>F_G = m\, a = m\, \frac{GM}{r^2}</math><br />
<br />
where ''m'' is still the ship's mass, ''G'' is the [[Template:G|gravitational constant]], ''M'' is the mass of the planet, and ''r'' is the distance from the ''center'' of the planet to the falling object.<br />
<br />
To find terminal velocity, we set ''F<sub>G</sub>'' equal to ''F<sub>D</sub>'':<br />
<br />
: <math>m\, \frac{GM}{r^2} = 0.5\, \rho\, v^2\, d\, (0.008 \cdot m)</math><br />
<br />
: <math>\frac{GM}{r^2} = 0.004\, \rho\, v^2\, d</math><br />
<br />
: <math>v = v_T = \sqrt{\frac{250\, GM}{r^2\, \rho\, d}}</math><br />
<br />
Assuming ''d'' is 0.2 (which is a good approximation, provided parachutes are not in use) and given that ''A'' is 1, this simplifies to:<br />
<br />
: <math>v_T = \sqrt{\frac{1250\, GM}{r^2\, \rho}}</math><br />
<br />
For the Mk16 pod and parachute example pictured above, the drag coefficient is 35.07, so its terminal velocity at sea level on Kerbin (which is 600 km from Kerbin's center) is:<br />
<br />
: <math>v_T = \sqrt{\frac{250\, GM}{r^2\, \rho \cdot 35.07}}</math><br />
<br />
: <math>\rho = 1.2230948554874 \cdot 1 \cdot e^{-0/5000}</math><br />
<br />
: <math>v_T = \sqrt{\frac{250 \cdot 6.674 \cdot 10^{-11} \cdot 5.2915793 \cdot 10^{22}}{600000^2 \cdot 1.2230948554874 \cdot 35.07}} = 7.56 \operatorname{m/s}</math><br />
<br />
=== Examples ===<br />
{| class="wikitable"<br />
!rowspan=2 valign=bottom| Altitude (m) ||colspan=6| v<sub>T</sub> (m/s)<br />
|-<br />
! Eve !! Kerbin !! Duna !! Jool !! Laythe<br />
|-<br />
| 0 || {{sigfigs|{{VT | planet=Eve | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 0}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 0}}|5}}<br />
|-<br />
| 100 || {{sigfigs|{{VT | planet=Eve | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 100}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 100}}|5}}<br />
|-<br />
| 1000 || {{sigfigs|{{VT | planet=Eve | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Duna | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Jool | alt= 1000}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt= 1000}}|5}}<br />
|-<br />
| 10000 || {{sigfigs|{{VT | planet=Eve | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Kerbin | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Duna | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Jool | alt=10000}}|5}} || {{sigfigs|{{VT | planet=Laythe | alt=10000}}|5}}<br />
|}<br />
<br />
== On-rails physics ==<br />
If a ship is "on rails" (meaning it's further than 2.25&nbsp;km from the actively-controlled ship) and its orbit passes through a planet's atmosphere, one of two things will happen based on atmospheric pressure at the ship's altitude:<br />
<br />
* below 0.01&nbsp;atm: no atmospheric drag will occur &mdash; the ship will be completely unaffected<br />
* 0.01&nbsp;atm or above: the ship will disappear<br />
<br />
The following table gives the altitude of this 0.01&nbsp;atm threshold for each celestial body with an atmosphere:<br />
{| class="wikitable"<br />
|-<br />
! Body || Altitude (m)<br />
|-<br />
| [[Eve]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Eve}}}}<br />
|-<br />
| [[Kerbin]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Kerbin}}}}<br />
|-<br />
| [[Duna]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Duna}}}}<br />
|-<br />
| [[Jool]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Jool}}}}<br />
|-<br />
| [[Laythe]] ||align="right"| {{Formatnum|{{PressureAltitude|pressure=0.01|body=Laythe}}}}<br />
|}<br />
<br />
== Atmospheric height ==<br />
The atmospheric height depend on the scale height of the celestial body and is where 0.000001<sup>th</sup> (0.0001&nbsp;%) of the surface pressure is remaining so the atmospheric pressure at the border isn't constant. Technically a craft in Jool's orbit can get lower into the atmosphere (or the atmosphere starts from a higher pressure).<br />
:<math>alt_{\text{atmospheric height}} = -ln\left(10^{-6}\right) \cdot \text{scale height}</math><br />
:<math>p_{\text{atmospheric height}} = p_0 \cdot 10^{-6}</math><br />
Kerbin's atmosphere ends at 0.000001&nbsp;atm and to calculate where the other celestial bodies should have the atmospheric height:<br />
:<math>alt_{\text{atmospheric height (real)}} = -ln\left(\frac{10^{-6}}{p_0}\right) \cdot \text{scale height}</math><br />
<br />
== Notes ==<br />
<references /></div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=API:FlightGlobals&diff=24765API:FlightGlobals2013-08-25T08:38:32Z<p>Tavert: Moved FlightGlobals.DragMultiplier from calculation of rho to calculation of dragForce, and added likely units for getAtmDensity and getStaticPressure (same units as assumed by MechJeb and Kerbal Engineer)</p>
<hr />
<div>The FlightGlobals class contains a number of useful static functions and fields.<br />
<br />
== Useful members ==<br />
<br />
{{Box|Warning.svg|This is not a complete listing of the class's members. It includes only the members that someone has figured out how to use and written an explanation for. If you figure out what the other members that aren't listed here do, you should add them to the list!|#FFDF80|Note}}<br />
<br />
A number of FlightGlobals functions have two versions: one that takes a CelestialBody as an argument and one that doesn't. Presumably the second version uses the mainBody of the active vessel. Below are listed only the versions that take an explicit CelestialBody argument.<br />
<br />
{| class="wikitable"<br />
! Signature<br />
! Description<br />
|-<br />
{{Method|returntype=float|name=GetAltitudeAtPos|args={{type|Vector3}} position, {{type|CelestialBody}} body|desc=Gives the altitude of a given world position above body's sea level.}}<br />
{{Method|returntype=double|name=getAtmDensity|args={{type|double}} pressure|desc=Calculates an atmospheric density in some arbitrary units (likely kg/m<sup>3</sup>). Presumably it's meant to take as input the output of FlightGlobals.getStaticPressure. The following code seems to work quite well to compute the drag force on a part and may in fact be the game's drag formula:<br />
<br />
<code>double rho = FlightGlobals.getAtmDensity(FlightGlobals.getStaticPressure(position, body));<br />
<br />
Vector3d airVel = vessel.orbit.GetVel() - body.getRFrmVel(position); //the velocity of the vessel relative to the air<br />
<br />
Vector3d dragForce = -0.5 * rho * FlightGlobals.DragMultiplier * part.mass * part.maximum_drag * airVel.magnitude * airVel;</code><br />
}}<br />
{{Method|returntype=Vector3d|name=getGeeForceAtPosition|args={{type|Vector3d}} pos|desc=The gravitational acceleration (not force) at the position pos. Does this compute which SOI pos is in or does it assume that pos is in the SOI of the active vessel's mainBody?}}<br />
{{Method|returntype=double|name=getStaticPressure|args={{type|Vector3d}} position, {{type|CelestialBody}} body|desc=The air pressure of body's atmosphere at position, in some arbitrary units (likely atmospheres).}}<br />
{{Property|type=Vessel|name=ActiveVessel|hasget=1|desc=The vessel currently being controlled by the player.}}<br />
{{Property|type=CelestialBody|array=[]|name=Bodies|hasget=1|desc=An array containing all the celestial bodies in the universe.}}<br />
{{Property|type=float|name=DragMultiplier|hasget=1|desc=A proportionality constant that scales the strength of aerodynamic drag forces.}}<br />
{{Property|type=List|array=&lt;{{type|Vessel}}&gt;|name=Vessels|hasget=1|desc=A list of all the Vessels in the universe.}}<br />
|}<br />
<br />
[[Category:Community API Documentation]]</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24672Talk:Atmosphere2013-08-24T11:22:22Z<p>Tavert: more comments on atmospheres</p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)<br />
:Maybe leave a note on [[User talk:Richardo 5|Richardo 5]]'s talk page. (S)He made “measurements” and then changed the constants. Di you measure a constant of 1.2002 at the height of the launch pad? Because I get a different value: e<sup>-68m/5000m</sup>×1.2230948554874=1.20657337<br />
:When I'm not mistaken, this would be the constant when you have 1 atm at the launch pad. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:06, 24 August 2013 (CDT)<br />
::It looks like it was [[User talk:Nulvinge|Nulvinge]] who added the ideal gas law equation but apparently without checking the numbers, then [[User talk:Richardo 5|Richardo 5]] plugged Nulvinge's ideal gas law numbers into a calculator and that's where the 1.2002 number came from. Here's a quick screenshot with MechJeb's density readout showing on the launch pad at 1.206 (close to your prediction), but it's kind of a short rocket so maybe on tall launch clamps you could get to 1.2002? http://i.imgur.com/FKz6PaY.jpg And part of my point was that you don't have 1 atm pressure at the launch pad, you have roughly 0.986 atm.<br />
::I think it should be noted rather than "the way the Kerbal universe works" that the 0.008 number comes from the API function FlightGlobals.DragMultiplier, and likewise with sea-level density from FlightGlobals.getAtmDensity(1.0), and these API functions are what mods like MechJeb and Kerbal Engineer use in their source to measure (and predict in MechJeb's case) drag and density. If these numbers from the API functions didn't reflect the game physics, then the atmospheric-landing and aerobraking predictions in MechJeb would have noticeably larger errors. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:39, 24 August 2013 (CDT)<br />
:::Ah nice. Yeah I guess this could be changed to the official numbers, maybe a little text that the density on the launch pad might differ. Sitting on the launch pad at about 68 m (AMSL) the measurement point should be about 26 meters above the launch pad (or 94 m AMSL). — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:23, 24 August 2013 (CDT)<br />
::::Having looked through the history and tracked it down to one user adding the ideal gas law without checking the numbers and another user replacing the numbers assuming the ideal gas law equation was reflecting KSP physics, I don't think the 1.2002 number, or discussion about the launch pad altitude, or the ideal gas law are necessary at all.<br />
::::I think it makes the most sense to say A = 0.008 (m^2/kg) * mass, and note that the 0.008 conversion factor comes from FlightGlobals.DragMultiplier. The density varies proportionally to pressure (with FlightGlobals.getAtmDensity(1.0) = 1.2230948554874), which is consistent with an isothermal atmosphere assumption so the temperature sensor readout is not at all related to pressure or density or drag, and the ideal gas law does not really hold for atmospheres in KSP.<br />
::::I could try going through and revising, but I'm not too familiar with wiki formatting or etiquette so I wanted to discuss first. It's a bit of a long article with numbers showing up in several places, any lingering inconsistencies from missed numbers would confuse people.<br />
::::A separate quibble that applies here and a few other places on the wiki, 1 atmosphere is defined by the CGPM as exactly 101325 Pascals. 101 kPa is a rounding, 101.327 kPa (which I see in a few places) is closer but also wrong. It's possible that Kerbal atmospheres means a different thing, but I think the original source of the 101.327 number is from this spreadsheet: https://docs.google.com/spreadsheet/ccc?key=0AuySrGPsDeq2dFdaS19xc2lobGc2aWNXUkJsZlVtWFE#gid=1 However I can't seem to find any mention of that number in the source forum thread or any of the linked pastebin data. FlightGlobals.getStaticPressure returns in atmospheres, and I guess the 0-altitude pressures for the bodies other than Kerbin were found from other sources (luckily they're all round numbers). Thanks for responding to this. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 06:22, 24 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24671Talk:Atmosphere2013-08-24T10:53:05Z<p>Tavert: </p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)<br />
:Maybe leave a note on [[User talk:Richardo 5|Richardo 5]]'s talk page. (S)He made “measurements” and then changed the constants. Di you measure a constant of 1.2002 at the height of the launch pad? Because I get a different value: e<sup>-68m/5000m</sup>×1.2230948554874=1.20657337<br />
:When I'm not mistaken, this would be the constant when you have 1 atm at the launch pad. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:06, 24 August 2013 (CDT)<br />
::It looks like it was [[User talk:Nulvinge|Nulvinge]] who added the ideal gas law equation but apparently without checking the numbers, then [[User talk:Richardo 5|Richardo 5]] plugged Nulvinge's ideal gas law numbers into a calculator and that's where the 1.2002 number came from. Here's a quick screenshot with MechJeb's density readout showing on the launch pad at 1.206 (close to your prediction), but it's kind of a short rocket so maybe on tall launch clamps you could get to 1.2002? http://i.imgur.com/FKz6PaY.jpg And part of my point was that you don't have 1 atm pressure at the launch pad, you have roughly 0.986 atm.<br />
::I think it should be noted rather than "the way the Kerbal universe works" that the 0.008 number comes from the API function FlightGlobals.DragMultiplier, and likewise with sea-level density from FlightGlobals.getAtmDensity(1.0), and these API functions are what mods like MechJeb and Kerbal Engineer use in their source to measure (and predict in MechJeb's case) drag and density. If these numbers from the API functions didn't reflect the game physics, then the atmospheric-landing and aerobraking predictions in MechJeb would have noticeably larger errors. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:39, 24 August 2013 (CDT)<br />
:::Ah nice. Yeah I guess this could be changed to the official numbers, maybe a little text that the density on the launch pad might differ. Sitting on the launch pad at about 68 m (AMSL) the measurement point should be about 26 meters above the launch pad (or 94 m AMSL). — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:23, 24 August 2013 (CDT)<br />
::::Having looked through the history and tracked it down to one user adding the ideal gas law without checking the numbers and another user replacing the numbers assuming the ideal gas law equation was reflecting KSP physics, I don't think the 1.2002 number, or discussion about the launch pad altitude, or the ideal gas law are necessary at all.<br />
::::I think it makes the most sense to say A = 0.008 (m^2/kg) * mass, and note that the 0.008 conversion factor comes from FlightGlobals.DragMultiplier. The density varies proportionally to pressure (with FlightGlobals.getAtmDensity(1.0) = 1.2230948554874), which is consistent with an isothermal atmosphere assumption so the temperature sensor readout is not at all related to pressure or density or drag, and the ideal gas law does not really hold for atmospheres in KSP.<br />
::::I could try going through and revising, but I'm not too familiar with wiki formatting or etiquette so I wanted to discuss first. It's a bit of a long article with numbers showing up in several places, any lingering inconsistencies from missed numbers would confuse people.<br />
::::A separate quibble that applies here and a few other places on the wiki, 1 atmosphere is defined by the CGPM as exactly 101325 Pascals. 101 kPa is a rounding, 101.327 kPa (which I see in a few places) is closer but also wrong. Thanks for responding to this. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 05:51, 24 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24670Talk:Atmosphere2013-08-24T10:51:06Z<p>Tavert: </p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)<br />
:Maybe leave a note on [[User talk:Richardo 5|Richardo 5]]'s talk page. (S)He made “measurements” and then changed the constants. Di you measure a constant of 1.2002 at the height of the launch pad? Because I get a different value: e<sup>-68m/5000m</sup>×1.2230948554874=1.20657337<br />
:When I'm not mistaken, this would be the constant when you have 1 atm at the launch pad. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:06, 24 August 2013 (CDT)<br />
::It looks like it was [[User talk:Nulvinge|Nulvinge]] who added the ideal gas law equation but apparently without checking the numbers, then [[User talk:Richardo 5|Richardo 5]] plugged Nulvinge's ideal gas law numbers into a calculator and that's where the 1.2002 number came from. Here's a quick screenshot with MechJeb's density readout showing on the launch pad at 1.206 (close to your prediction), but it's kind of a short rocket so maybe on tall launch clamps you could get to 1.2002? http://i.imgur.com/FKz6PaY.jpg And part of my point was that you don't have 1 atm pressure at the launch pad, you have roughly 0.986 atm.<br />
::I think it should be noted rather than "the way the Kerbal universe works" that the 0.008 number comes from the API function FlightGlobals.DragMultiplier, and likewise with sea-level density from FlightGlobals.getAtmDensity(1.0), and these API functions are what mods like MechJeb and Kerbal Engineer use in their source to measure (and predict in MechJeb's case) drag and density. If these numbers from the API functions didn't reflect the game physics, then the atmospheric-landing and aerobraking predictions in MechJeb would have noticeably larger errors. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:39, 24 August 2013 (CDT)<br />
:::Ah nice. Yeah I guess this could be changed to the official numbers, maybe a little text that the density on the launch pad might differ. Sitting on the launch pad at about 68 m (AMSL) the measurement point should be about 26 meters above the launch pad (or 94 m AMSL). — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:23, 24 August 2013 (CDT)<br />
::::Having looked through the history and tracked it down to one user adding the ideal gas law without checking the numbers and another user replacing the numbers assuming the ideal gas law equation was reflecting KSP physics, I don't think the 1.2002 number, or discussion about the launch pad altitude, or the ideal gas law are necessary at all.<br />
::::I think it makes the most sense to say A = 0.008 (m^2/kg) * m, and note that the 0.008 conversion factor comes from FlightGlobals.DragMultiplier. The density varies proportionally to pressure, which is consistent with an isothermal atmosphere assumption so the temperature sensor readout is not at all related to pressure or density or drag, and the ideal gas law does not really hold for atmospheres in KSP.<br />
::::I could try going through and revising, but I'm not too familiar with wiki formatting or etiquette so I wanted to discuss first. It's a bit of a long article with numbers showing up in several places, any lingering inconsistencies from missed numbers would confuse people.<br />
::::A separate quibble that applies here and a few other places on the wiki, 1 atmosphere is defined by the CGPM as exactly 101325 Pascals. 101 kPa is a rounding, 101.327 kPa (which I see in a few places) is closer but also wrong. Thanks for responding to this. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 05:51, 24 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24661Talk:Atmosphere2013-08-24T09:44:18Z<p>Tavert: </p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)<br />
:Maybe leave a note on [[User talk:Richardo 5|Richardo 5]]'s talk page. (S)He made “measurements” and then changed the constants. Di you measure a constant of 1.2002 at the height of the launch pad? Because I get a different value: e<sup>-68m/5000m</sup>×1.2230948554874=1.20657337<br />
:When I'm not mistaken, this would be the constant when you have 1 atm at the launch pad. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:06, 24 August 2013 (CDT)<br />
::It looks like it was [[User talk:Nulvinge|Nulvinge]] who added the ideal gas law equation but apparently without checking the numbers, then [[User talk:Richardo 5|Richardo 5]] plugged Nulvinge's ideal gas law numbers into a calculator and that's where the 1.2002 number came from. Here's a quick screenshot with MechJeb's density readout showing on the launch pad at 1.206 (close to your prediction), but it's kind of a short rocket so maybe on tall launch clamps you could get to 1.2002? http://i.imgur.com/FKz6PaY.jpg And part of my point was that you don't have 1 atm pressure at the launch pad, you have roughly 0.986 atm.<br />
::I think it should be noted rather than "the way the Kerbal universe works" that the 0.008 number comes from the API function FlightGlobals.DragMultiplier, and likewise with sea-level density from FlightGlobals.getAtmDensity(1.0), and these API functions are what mods like MechJeb and Kerbal Engineer use in their source to measure (and predict in MechJeb's case) drag and density. If these numbers from the API functions didn't reflect the game physics, then the atmospheric-landing and aerobraking predictions in MechJeb would have noticeably larger errors. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:39, 24 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24660Talk:Atmosphere2013-08-24T09:40:47Z<p>Tavert: </p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)<br />
:Maybe leave a note on [[User talk:Richardo 5|Richardo 5]]'s talk page. (S)He made “measurements” and then changed the constants. Di you measure a constant of 1.2002 at the height of the launch pad? Because I get a different value: e<sup>-68m/5000m</sup>×1.2230948554874=1.20657337<br />
:When I'm not mistaken, this would be the constant when you have 1 atm at the launch pad. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:06, 24 August 2013 (CDT)<br />
::It looks like it was [[User talk:Nulvinge|Nulvinge]] who added the ideal gas law equation but apparently without checking the numbers, then [[User talk:Richardo 5|Richardo 5]] plugged Nulvinge's ideal gas law numbers into a calculator and that's where the 1.2002 number came from. Here's a quick screenshot with MechJeb's density readout showing on the launch pad at 1.206 (close to your prediction), but it's kind of a short rocket so maybe on tall launch clamps you could get to 1.2002? http://i.imgur.com/FKz6PaY.jpg And part of my point was that you don't have 1 atm pressure at the launch pad, you have roughly 0.986 atm.<br />
::I think it should be noted rather than "the way the Kerbal universe works" that this number comes from the API function FlightGlobals.DragMultiplier, and likewise with sea-level density from FlightGlobals.getAtmDensity(1.0), and these API functions are what mods like MechJeb and Kerbal Engineer use in their source to measure (and predict in MechJeb's case) drag and density. If these numbers from the API functions didn't reflect the game physics, then the atmospheric-landing and aerobraking predictions in MechJeb would have noticeably larger errors. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:39, 24 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24659Talk:Atmosphere2013-08-24T09:39:48Z<p>Tavert: more density discussion</p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)<br />
:Maybe leave a note on [[User talk:Richardo 5|Richardo 5]]'s talk page. (S)He made “measurements” and then changed the constants. Di you measure a constant of 1.2002 at the height of the launch pad? Because I get a different value: e<sup>-68m/5000m</sup>×1.2230948554874=1.20657337<br />
:When I'm not mistaken, this would be the constant when you have 1 atm at the launch pad. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:06, 24 August 2013 (CDT)<br />
::It looks like it was [[User talk:Nulvinge|Nulvinge]] who added the ideal gas law equation but apparently without checking the numbers, then [[User talk:Richardo 5|Richardo 5]] plugged Nulvinge's ideal gas law numbers into a calculator and that's where the 1.2002 number came from. Here's a quick screenshot with MechJeb's density readout showing on the launch pad at 1.206 (close to your prediction), but it's kind of a short rocket so maybe on tall launch clamps you could get to 1.2002? http://i.imgur.com/FKz6PaY.jpg<br />
::I think it should be noted rather than "the way the Kerbal universe works" that this number comes from the API function FlightGlobals.DragMultiplier, and likewise with sea-level density from FlightGlobals.getAtmDensity(1.0), and these API functions are what mods like MechJeb and Kerbal Engineer use in their source to measure (and predict in MechJeb's case) drag and density. If these numbers from the API functions didn't reflect the game physics, then the atmospheric-landing and aerobraking predictions in MechJeb would have noticeably larger errors. --[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 04:39, 24 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24651Talk:Atmosphere2013-08-24T00:09:29Z<p>Tavert: /* Density */</p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level density number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24650Talk:Atmosphere2013-08-24T00:08:54Z<p>Tavert: /* Density */</p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
<br />
Notice how we've been using Elembis' original template http://wiki.kerbalspaceprogram.com/wiki/Template:VT with the correct 1.223 sea-level denisty number to calculate terminal velocity in every article, which has been giving correct terminal velocity values. But he referenced this more-commonly-seen Atmosphere article as his source, without realizing his correct numbers would gradually be replaced by incorrect numbers.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)</div>Taverthttps://wiki.kerbalspaceprogram.com/index.php?title=Talk:Atmosphere&diff=24649Talk:Atmosphere2013-08-23T23:59:37Z<p>Tavert: discussion about density</p>
<hr />
<div>== Drag calculation notes ==<br />
<br />
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.<br />
<br />
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.<br />
<br />
<br />
Edit:<br />
<br />
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 15:48, 15 July 2013 (CDT)<br />
<br />
<br />
:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 16:26, 15 July 2013 (CDT)<br />
<br />
::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] ([[User talk:T0w0i7ne|talk]]) 22:39, 15 July 2013 (CDT)<br />
:::Now about the physical simulation, I have no idea. Maybe this should be asked on the IRC. As I don't know how they calculate the crosssection of the craft, they maybe tried to make that calculation easier and don't calculate the cross section as precise as it needed and instead they estimate a heavy craft has a bigger cross section. I don't know ;). About etiquette: In a forum you can see who posted a post, but here it is harder without the signature. And it is always nice to know who you are talking to. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 05:19, 16 July 2013 (CDT)<br />
:::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT)<br />
::::This is exactly what I was trying to get at in my first post. I understand the physics/aerodynamics. I'm confused by the completely unrealistic way that drag is supposedly being modeled in the game (see the wiki page). When I said "mass specific forces/accelerations" I meant it in the context that the two concepts are identical ( i.e. for all quantities being instantaneous F/m = a ). So to calculate the acceleration of the craft due to the force of drag one would divide the force of drag by the craft's mass (F_d = 1/2 * rho * A * d * V^2 and a_d = F_d/m). So the equation on the wiki page (F_d = 1/2 * rho * A * d * m * V^2) provides a meaningless value as far as I can tell. I just do not understand the reasoning behind the decision to use such a meaningless value to model drag in the game (once again assuming that the statement in the wiki that the game does it this way is accurate). — [[User:T0w0i7ne|T0w0i7ne]] <small>&#91;[[User talk:T0w0i7ne|talk]]&#93;</small> 00:47, 17 July 2013 (MDT)<br />
:::::Yes the way it is calculated in the stock game is completely unrealistic. This is because mass substituted area in the equation, among many other things. For more realistic drag, FAR is available as a MOD. It's primarily used to make planes behave more realistically, but it replaces the drag and lift algorithms to achieve this, thus making drag and lift significantly more realistic. (It makes them about the same as something like Microsoft Flight Simulator. Still far less realistic than Flight Gear or X-Plane. Possibly also on par with many of the R/C Flight Simulators out there.)--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 02:10, 17 July 2013 (CDT)<br />
::::Ehm Ruedii, why [http://wiki.kerbalspaceprogram.com/w/index.php?title=Talk:Atmosphere&diff=21959&oldid=21922 first edit] here was a little failed (removed my comment and readded the old signatures of T0w0i7ne). I restored it with this edit. — [[User:XZise|xZise]] <small>&#91;[[User talk:XZise|talk]]&#93;</small> 04:00, 17 July 2013 (CDT)<br />
<br />
== Density ==<br />
<br />
The ideal gas law is irrelevant for calculating density. Real-world physics do not always apply to Kerbal atmospheres. Elembis' original number of 1.2230948554874 was correct. Notice the comment he included, density at sea level is given by FlightGlobals.getAtmDensity(1.0). For p = 1.0 atm, the 1.223 number is more accurate, see http://i.imgur.com/wJPO4m2.png for verification. The 1.2002 that this page has been quoting might be accurate at the altitude of the launch pad, where pressure is slightly lower than 1 atm.<br />
<br />
Similarly, the factor of 0.008 isn't on the density. The number comes from FlightGlobals.DragMultiplier, and it should either be interpreted as the ratio of cross-sectional area in m^2 per kilogram of mass for KSP parts, or as a dimensionless factor you have to multiply the drag coefficients (usually 0.2) by.<br />
--[[User:Tavert|Tavert]] ([[User talk:Tavert|talk]]) 18:59, 23 August 2013 (CDT)</div>Tavert