Re: IPv6 Burgers (was: IPv6 Ignorance)
Another measure of the size of the IPv6 address space... Back on World IPv6 Day in June 2011, Dartware had a barbecue. (Why? Because the burgers had 128 (bacon) bits and we served IP(A) to drink :-) You can see some photos at: http://www.networkworld.com/community/blog/scenes-ipv6-day-barbecue But we came up with another interesting measure for the vastness of the IPv6 address space: If an IPv4 hamburger patty has 2^32 (4.2 billion) unique addresses in its 1/4 inch thickness, how thick would an IPv6 hamburger be (with 2^128 unique addresses)? The answer is... 53 billion light-years. It's straightforward unit conversions. There are 2^96 IPv4 Hamburgers at a quarter-inch apiece. That's 2^96 inches/4 (2^94 inches). Switching to decimal units, 1.98x10^32 inches; 1.65x10^27 feet; 3.13x10^23 miles; and then continuing to convert to light-years. A good tool for this kind of wacky unit conversion is Frink (http://futureboy.us/fsp/frink.fsp?fromVal=2%5E94+inches&toVal=lightyears), which can do this in one shot. Simply enter: From: 2^94 inches To: lightyears and you'll see the answer! Rich Brown richard.e.brown@dartware.com Dartware, LLC http://www.intermapper.com 66-7 Benning Street Telephone: 603-643-9600 West Lebanon, NH 03784-3407 Fax: 603-643-2289
On 2012-09-17 13:48, Richard Brown wrote:
Another measure of the size of the IPv6 address space... Back on World IPv6 Day in June 2011, Dartware had a barbecue. (Why? Because the burgers had 128 (bacon) bits and we served IP(A) to drink :-) You can see some photos at: http://www.networkworld.com/community/blog/scenes-ipv6-day-barbecue
But we came up with another interesting measure for the vastness of the IPv6 address space:
If an IPv4 hamburger patty has 2^32 (4.2 billion) unique addresses in its 1/4 inch thickness, how thick would an IPv6 hamburger be (with 2^128 unique addresses)?
The answer is... 53 billion light-years.
Just got to playing with this today, trying to put it in some sort of perspective. First off, lets bring that down to human-sized numbers, using standard units used in astronomy: 2^94 inches = 16 gigaparsecs + 304 megaparsecs + 322 kiloparsecs + 752 parsecs + 2 lightyears + 57101 au + 23233 earthradius (Gigaparsecs isn't very common, but that's because it's a bit big.) So... How big is that? What can we compare it to? Well, let's start at the top: does this thing actually fit in our universe? The size of the observable universe is set by the Hubble Constant and lightspeed: The Hubble Constant is the rate of growth of expansion in the universe - the redshift phenomena. The further away you look, the faster things are moving away from us. At a certain point, they are moving away from us faster than light, meaning that light coming from them would never reach us. That's about 14 gigaparsecs away. (Adjusting for such things as how much they will have moved since you measured them. There's a whole rabbit hole to go down for this, on Wikipedia alone.) Which means the observable universe is about 28 gigaprsecs across. (Now you can see why gigaparsecs isn't a common unit.) So our hamburger patty would fit inside it - but you wouldn't be able to see one end from the other. Ever. In fact, while someone at the center could reach either end, once they got there they'd never be able to reach the other. They wouldn't even be able to get back to where they started. Which of course means that even if you ate at lightspeed, you'd never be able to eat it. (Oh, and if it still has a radius of 3 inches - standard 1/4 pound burger at 1/4 inch thick - it's got a volume around that of 11,000 Earths, and a mass of about 1,400 Earths, about 4.6 times the mass of Jupiter.)
It's straightforward unit conversions. There are 2^96 IPv4 Hamburgers at a quarter-inch apiece. That's 2^96 inches/4 (2^94 inches). Switching to decimal units, 1.98x10^32 inches; 1.65x10^27 feet; 3.13x10^23 miles; and then continuing to convert to light-years.
A good tool for this kind of wacky unit conversion is Frink
(http://futureboy.us/fsp/frink.fsp?fromVal=2%5E94+inches&toVal=lightyears), which can do this in one shot. Simply enter:
I prefer the 'units' program, which is usually a standard utility on Unix-like boxes. (If not in your distro of choice, finding the GNU or BSD versions is left as an exercise for the reader. ;) ) Daniel T. Staal --------------------------------------------------------------- This email copyright the author. Unless otherwise noted, you are expressly allowed to retransmit, quote, or otherwise use the contents for non-commercial purposes. This copyright will expire 5 years after the author's death, or in 30 years, whichever is longer, unless such a period is in excess of local copyright law. ---------------------------------------------------------------
On Mon, Sep 17, 2012 at 1:48 PM, Richard Brown <richard.e.brown@dartware.com> wrote:
Another measure of the size of the IPv6 address space... Back on World IPv6 Day in June 2011, Dartware had a barbecue. (Why? Because the burgers had 128 (bacon) bits and we served IP(A) to drink :-) You can see some photos at: http://www.networkworld.com/community/blog/scenes-ipv6-day-barbecue
But we came up with another interesting measure for the vastness of the IPv6 address space:
If an IPv4 hamburger patty has 2^32 (4.2 billion) unique addresses in its 1/4 inch thickness, how thick would an IPv6 hamburger be (with 2^128 unique addresses)?
The answer is... 53 billion light-years.
Interesting. If a teeny dot of gristle at the edge of the patty is an IPv4 /28 LAN and the same LAN is /64 in IPv6, how big is IPv6 now? The 1/4 inch patty holds all the IPv4 LANs whose IPv4 capacity I'm calling 2^28. IPv6's in principle has 2^64 LANs, so an increase of 2^36 LANs. Patty was 1/4 inch, so our IPv6 patty is 2^32 inches or about 10 billionths of a lightyear. 68,000 miles, a little over a quarter of the distance to the moon. That's a big burger. But not so big as you thought. By 17 orders of magnitude. In point of fact, a mere 150,000 people put together will eat all that beef in their lifetimes. But for the distribution problem, the world population could chew through it in half a day. Worse, that's if we were managing IPv6 delegations the way we manage IPv4 delegations. We're not. We're using sparse allocation. And 6RD. And default customer allocations of 65,000 LANs. And other interesting stuff that drastically increases the consumption characteristics. Nice burger. Om nom nom nom. Regards, Bill Herrin -- William D. Herrin ................ herrin@dirtside.com bill@herrin.us 3005 Crane Dr. ...................... Web: <http://bill.herrin.us/> Falls Church, VA 22042-3004
On Thu, Sep 20, 2012 at 4:29 PM, William Herrin <bill@herrin.us> wrote:
The 1/4 inch patty holds all the IPv4 LANs whose IPv4 capacity I'm calling 2^28. IPv6's in principle has 2^64 LANs, so an increase of 2^36 LANs. Patty was 1/4 inch, so our IPv6 patty is 2^32 inches or about 10 billionths of a lightyear. 68,000 miles, a little over a quarter of the distance to the moon.
2^34, my bad. So you get a full moon burger out of it. -Bill -- William D. Herrin ................ herrin@dirtside.com bill@herrin.us 3005 Crane Dr. ...................... Web: <http://bill.herrin.us/> Falls Church, VA 22042-3004
In message <CAP-guGVaKSokeQEVX7aysEYQrr7g2ErREAAsXNwAyabQpqM5jw@mail.gmail.com>, W illiam Herrin writes:
Worse, that's if we were managing IPv6 delegations the way we manage IPv4 delegations. We're not. We're using sparse allocation. And 6RD. And default customer allocations of 65,000 LANs. And other interesting stuff that drastically increases the consumption characteristics.
6rd can be dense as native deployment. In fact it is not hard to do so and if you are using the shared /10 just allocated or RFC 1918 between you and your customers more than once simple map all of IPv4 space does not work. RFC 5969 does NOT say you must use 32 bits and actually lots of examples where it isn't done.
Nice burger. Om nom nom nom.
Regards, Bill Herrin
--=20 William D. Herrin ................ herrin@dirtside.com bill@herrin.us 3005 Crane Dr. ...................... Web: <http://bill.herrin.us/> Falls Church, VA 22042-3004
-- Mark Andrews, ISC 1 Seymour St., Dundas Valley, NSW 2117, Australia PHONE: +61 2 9871 4742 INTERNET: marka@isc.org
On Thu, Sep 20, 2012 at 7:50 PM, Mark Andrews <marka@isc.org> wrote:
In message <CAP-guGVaKSokeQEVX7aysEYQrr7g2ErREAAsXNwAyabQpqM5jw@mail.gmail.com>, W illiam Herrin writes:
Worse, that's if we were managing IPv6 delegations the way we manage IPv4 delegations. We're not. We're using sparse allocation. And 6RD. And default customer allocations of 65,000 LANs. And other interesting stuff that drastically increases the consumption characteristics.
6rd can be dense as native deployment. In fact it is not hard to
Have you heard the one about the difference between theory and practice? Regards, Bill Herrin -- William D. Herrin ................ herrin@dirtside.com bill@herrin.us 3005 Crane Dr. ...................... Web: <http://bill.herrin.us/> Falls Church, VA 22042-3004
participants (4)
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Daniel Staal -
Mark Andrews -
Richard Brown -
William Herrin