Difference between revisions of "User:XZise/Atmospheric Height"

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Hey, I stumbled upon this one, when I worked on the [[Template:Infobox/Body|Infobox/Body]] template at the Wiki.
 
Hey, I stumbled upon this one, when I worked on the [[Template:Infobox/Body|Infobox/Body]] template at the Wiki.
  
According to the [[Atmosphere#Drag|wiki]], the pressure at a specific height could be calculated by:
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=== Scale height ===
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According to the [[Atmosphere#Drag|wiki]] and [[w:Scale height|wikipedia], the pressure at a specific height could be calculated by:
  
 
:p<SUB>alt</SUB> = p<SUB>0</SUB> × e<SUP>-alt/H</SUP>
 
:p<SUB>alt</SUB> = p<SUB>0</SUB> × e<SUP>-alt/H</SUP>
  
Now the atmospheric height (in the template) is calculated by:
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=== Atmosphere height ===
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Atmospheric height is defined as the altitude where pressure falls bellow a millionth the pressure at sea level:
  
 
:alt<SUB>atmo</SUB> = -ln(10⁻⁶)×H
 
:alt<SUB>atmo</SUB> = -ln(10⁻⁶)×H
  
Now H is the scale height which could be found on the wiki but not important for the next step. I want to know the pressure at the atmospheric height. Now in theory this should be constant, as the drag also depend on the pressure (that's why you don't need to run any engine in space to maintain the velocity). So lets calculate this pressure:
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To check this, compute the pressure at altitude alt<SUB>atmo</SUB>:
  
 
:{|
 
:{|
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|}
 
|}
  
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It checks out. Kerbin this has 1 atm at sea level and 10⁻⁶ atm at atmospheric height; Jool has 15 atm at sea level and 15×10⁻⁶ atm left at atmospheric height.
  
Woooo so, the pressure at the atmospheric height is simply <I>p_0 × 10⁻⁶</I> where p_0 is the pressure at sea level. On Kerbin this is 1 atm on sea level and 10⁻⁶ atm at the atmospheric height. But Jool for example have still 15×10⁻⁶ atm left at the atmospheric height, as the pressure on “sea level” is 15 atm. Now doing the reverse: At which altitude is 15×10⁻⁶ atm on Kerbin. By my understanding, this should be a save height as crafts in Jool's orbit won't be slowed down at this height/pressure.
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=== Deriving altitude from pressure ===
  
Now to calculate the altitude for a specific pressure change the formula:
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Now, suppose we want to find where on Kerbin pressure is 15×10⁻⁶ atm. We simply reverse the scale height formula:
  
 
:{|
 
:{|
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|}
 
|}
  
If you look closely you can see it looks like the “atmospheric height” formula, with the difference that p<sub>alt</sub>/p<sub>0</sub> is 10⁻⁶. So now for Kerbin we use for p<SUB>alt</SUB> (the pressure for the searched altitude) is 15×10⁻⁶ atm, p<SUB>0</SUB> is 1 atm and H is 5000 m. This is now saying 55537.3018 meters.
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This is very similar to the expression for alt<SUB>atmo</SUB>. It makes sense since, in both case, we compute an altitude for a given pressure (absolute or relative to pressure at sea level).
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Plugging in the values p<SUB>alt</SUB> = 15×10⁻⁶ atm, p<SUB>0</SUB> = 1 atm and H = 5000 m, it gives us the 55537.3018 meters above sea level.
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=== Safe altitude ===
  
So applying Jool's “atmospheric height pressure” on Kerbin means, that you can savely orbit at about 56 km which is way lower than 69 km. On the other side, the altitude of 10⁻⁶ atm at Jool is 165 235.608 m so about 30 km higher. Eve has a difference of 10 km and Duna about -5 km (as p<SUB>0</SUB> is lower than 1 atm).
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Since there is no drag effect on Jool right above atmospheric height with p = 15×10⁻⁶ atm, we could expect an orbit in Kerbin's atmosphere at this altitude (i.e. 56km) to be safe as well.
  
Now you maybe saying that 15×10⁻⁶ atm and 10⁻⁶ atm aren't very far away (although the different altitudes are pretty far away as 56 km are only 80 % of 69 km) you know could construct an example with an sea level pressure of 10⁺⁶ atm resulting in a pressure of 1 atm at the “atmospheric height” which is the pressure at Kerbin's sea level.
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Note: since the threshold defining atmosphere height is relative to pressure at sea level, the pressure at atmosphere height is not guaranteed to actually be negligible
  
So either I mixed something up or the calculation for the atmospheric height is wrong or the formula for the atmospheric drag in the wiki is wrong.
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Kerbal Space Program uses a simplified model to allow logical acceleration by ignoring atmospheric drag above an arbitrary limit. In a more precise model of Kerbin, vessels would still experience orbital decay above 69km due to atmospheric drag. Since drag is actually taken into account below 69km, any orbit whose periapsis is below this altitude will ultimately decay in the game; however, even at 50km this would take a really long time (only physical ×4 acceleration would be available).
  
So please let me know who does what wrong. I don't want to file a bug report as I'm not sure if the problem is on my end.
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Finally, there is another trick to this: drag is not simulated at all on background vessels (e.g. dropped stages farther than 2.5km). Instead, the vessel can orbit safely even in the atmosphere (unless switched to), but is considered as having crashed if its altitude falls below some threshold It looks like this limit is around ~30km on Kerbin, but more Science © is needed to derive the way this is computed.

Revision as of 09:20, 26 December 2014

Hey, I stumbled upon this one, when I worked on the Infobox/Body template at the Wiki.

Scale height

According to the wiki and [[w:Scale height|wikipedia], the pressure at a specific height could be calculated by:

palt = p0 × e-alt/H

Atmosphere height

Atmospheric height is defined as the altitude where pressure falls bellow a millionth the pressure at sea level:

altatmo = -ln(10⁻⁶)×H

To check this, compute the pressure at altitude altatmo:

patmo = p0 × e-altatmo/H
= p0 × e-(-ln(10⁻⁶)×H)/H
= p0 × eln(10⁻⁶)
= p0 × 10⁻⁶

It checks out. Kerbin this has 1 atm at sea level and 10⁻⁶ atm at atmospheric height; Jool has 15 atm at sea level and 15×10⁻⁶ atm left at atmospheric height.

Deriving altitude from pressure

Now, suppose we want to find where on Kerbin pressure is 15×10⁻⁶ atm. We simply reverse the scale height formula:

palt = p0 × e-alt/H | ÷p0
palt/p0 = e-alt/H | ln()
ln(palt/p0) = -alt/H | ×-H
-ln(palt/p0) × H = alt

This is very similar to the expression for altatmo. It makes sense since, in both case, we compute an altitude for a given pressure (absolute or relative to pressure at sea level).

Plugging in the values palt = 15×10⁻⁶ atm, p0 = 1 atm and H = 5000 m, it gives us the 55537.3018 meters above sea level.

Safe altitude

Since there is no drag effect on Jool right above atmospheric height with p = 15×10⁻⁶ atm, we could expect an orbit in Kerbin's atmosphere at this altitude (i.e. 56km) to be safe as well.

Note: since the threshold defining atmosphere height is relative to pressure at sea level, the pressure at atmosphere height is not guaranteed to actually be negligible

Kerbal Space Program uses a simplified model to allow logical acceleration by ignoring atmospheric drag above an arbitrary limit. In a more precise model of Kerbin, vessels would still experience orbital decay above 69km due to atmospheric drag. Since drag is actually taken into account below 69km, any orbit whose periapsis is below this altitude will ultimately decay in the game; however, even at 50km this would take a really long time (only physical ×4 acceleration would be available).

Finally, there is another trick to this: drag is not simulated at all on background vessels (e.g. dropped stages farther than 2.5km). Instead, the vessel can orbit safely even in the atmosphere (unless switched to), but is considered as having crashed if its altitude falls below some threshold It looks like this limit is around ~30km on Kerbin, but more Science © is needed to derive the way this is computed.