Last time we have come to the conclusion that, indeed, the air at the top of the Dome can be significantly colder than that near the surface.

Provided, of course, we consider the Dome to be permeable by air. I can assume, that if this whole system is closed and does not interact with the rest of the atmosphere in any way, its temperature gradient might behave differently. And there is a strong evidence that this invisible barrier is impenetrable, because once it has been hit by a MOAB. And nothing bad happened to its inhabitants.

Hmm, how can I be sure? If only this barrier wasn’t some unexplained, mysterious phenomenon, that cannot be described. Oh, wait.

For the time being, let’s assume there’s some interaction and flow between inside and outside of the Dome (at least on particle scale or microscopic level). And the air at the top has the temperature of roughly -30 degree Celsius as calculated before.

Is there really enough cold air (in proportion to warmer, near the surface) to cause freezing when mixed up? And what is cold air and what, exactly is warm?

If the temperature at ground level is 21°C, then at 1000 m it is 14.5°C, at 2 km it equals 8°C, at 3 km it is 1.5°C and at about little more than 3.23 km it drops below the value of freezing point for water.

I will use this as my divide boundary. What is below is ‘**Warm Air**‘ from now on, what is above this 3.23 km level I will consider as ‘**Cold**‘.

What is the ratio between those two? Well, if it was a cube or rectangular hexahedron, not a dome, we would say that there’s about 40% of Warm Air and the rest is Cold Air. Case solved, move on.

But a dome is not a cube. Its volume is not s^{3} (s is length of any side). It equals two thirds of its radius cubed and multiplied by π.

I will calculate the **volume of the whole Dome**. Radius is 8.045km, so:

**V = 2/3 * 8.045 ^{3} * 3.1416 = 2/3 * 520.689 * 3.1416 = 2/3 * 1635.7966 = 1090.5311 km^{3} ^{
}**

Now, we need to somehow find out the radius of the smaller circle that is created by sectioning the Dome with a plane parallel to its base at the height of 3.23 km. Actually, I sort of cheated here and constructed a semicircle in Inkscape (with grid and exact values in centimeters) and by using precisely measured rectangles I was able to estimate this radius to be around 7.373 km. Correct me if I am wrong here, please!

The** Volume of Cold Air **needs a slightly modified equation: It’s no longer a hemisphere, it is a spherical cap, i.e. its height is lower than the radius of its base. Radius is 7.373 km, height is (8.045 km – 3.23 km) 4.815 km.

**V = 1/6 * π * height (3 * radius ^{2} + height^{2})**

**V = 420.11 ****km ^{3} **

**Volume of Warm Air** is thus: 1090.53 – 420.11 = 670.42 km^{3} ^{
}

So, inside the Dome, actually 61% of air has temperature above 0°C.

It does not look like this because picture is not to scale. I will work on this, I promise.

*I have to stop here, I need to see a doctor. To be continued.*