3 ms·
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 uni
by throw0101a 5y ago
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 https://www.youtube.com/watch?v=7Tnh4upTOC4
- deleted 5y ago[deleted]
- JuettnerDistrib 5y agoPerhaps 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 5y ago> 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 5y agoFull surface including oceans. I actually got something like 6.6 * 10^17 per mm^2, but who's counting?