[throw0101a]

Another way of thinking about it:

* Stars in the Milky Way: 400 Billion

* Galaxes in the universe: 2 Trillion

So (4x10^11)x(2x10^12)=8x10^23 stars in the universe.

* Size of IPv6 address space: 3.4x10^38

Find the ratio between addresses and stars:

* 3.4x10^38 / 8x10^23

IPv6 offers about 430 trillion times more addresses than estimated stars in the universe.

From Tom Coffee's presentation "An Enterprise IPv6 Address Planning Case-Study"

* https://www.youtube.com/watch?v=7Tnh4upTOC4

[JuettnerDistrib]

Perhaps in more human terms:

On the surface of the Earth, there are 8.4 IPv4 addresses per km^2. Not counting the oceans, that would be 28 IPv4 addresses per km^2 land.

IPv6 gives 10^17 addresses per mm^2 (yes, square millimeter).

In terms of volume, 10^8 IPv6 addresses per mm^3 throughout the Earth.

[throw0101a]

> IPv6 gives 10^17 addresses per mm^2 (yes, square millimeter).

Not that it practically matters, but: is that the 'full surface' or not counting the oceans (land-only)?

[JuettnerDistrib]

Full surface including oceans. I actually got something like 6.6 * 10^17 per mm^2, but who's counting?