User:XZise/Atmospheric Height

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Revision as of 12:56, 19 July 2013 by XZise (talk | contribs)
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Hey, I stumbled upon this one, when I worked on the Infobox/Body template at the Wiki.

According to the wiki, the pressure at a specific height could be calculated by:

palt = p0 × e-alt/H

Now the atmospheric height (in the template) is calculated by:

altatmo = -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:

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


Woooo so, the pressure at the atmospheric height is simply p_0 × 10⁻⁶ 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.

Now to calculate the altitude for a specific pressure change the formula:

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

If you look closely you can see it looks like the “atmospheric height” formula, with the difference that palt/p0 is 10⁻⁶. So now for Kerbin we use for palt (the pressure for the searched altitude) is 15×10⁻⁶ atm, p0 is 1 atm and H is 5000 m. This is now saying 55537.3018 meters.

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 p0 is lower than 1 atm).

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.

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.

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.