Pigeons' Bandwidth Advantage Quantified 462
An anonymous reader submits "A well documented test took
place in the north of Israel, in presence of several dozen Internet geeks and
experts. During the test, 3 homing pigeons carried 4 GB (gigabytes) for 100 km
distance, achieving, what apparently looks as pigeons' world record in data
transfer to a given distance. Bandwidth achieved by the pigeons was 2.27
Mbps...Transferring a similar volume of information through a common uplink of
ADSL line would have taken no less than 96 hours..."
Re:Sure (Score:3, Informative)
latency v. bandwidth (Score:4, Informative)
Re:Back of envalope (Score:4, Informative)
Anm
Re:It begins... (Score:3, Informative)
Is it already April 1st somewhere?
That may well be the case, but stranger things [eagle.auc.ca] have happened.
Re:The lag will be a problem, though... (Score:3, Informative)
Re:One of those things that shouldn't surprise but (Score:5, Informative)
Re:Ha! (Score:5, Informative)
A little more "Birdseed for Thought" (Score:4, Informative)
This FAQ [interbug.com] answered that question and many others for me.
Re:latency v. bandwidth (Score:5, Informative)
"The numerical difference between the upper and lower frequencies of a band of electromagnetic radiation, especially an assigned range of radio frequencies." (thank you Google).
And under that definition, these pigeons have no bandwidth (unless you're counting the frequency at which they flap their wings
The Jargon File [catb.org] says
"Used by hackers (in a generalization of its technical meaning) as the volume of information per unit time that a computer, person, or transmission medium can handle. "Those are amazing graphics, but I missed some of the detail -- not enough bandwidth, I guess." Compare low-bandwidth. This generalized usage began to go mainstream after the Internet population explosion of 1993-1994. 2. Attention span. 3. On Usenet, a measure of network capacity that is often wasted by people complaining about how items posted by others are a waste of bandwidth."
Re:latency v. bandwidth (Score:3, Informative)
Incorrect. The bandwidth would remain constant - what would double if the distance doubled is what is usually called a "bandwidth-delay product". This quantity represents the pipe capacity of a given length, or in other words the amount of data in transit at any given moment. This is of course assuming you have an unlimited number of pigeons you can keep sending out.
You would be correct, using the term bandwidth loosely, if the number of pigeons stayed constant. However, using the strict definition, bandwidth is totally unrelated to line latency/round trip time.
but seriously (Score:4, Informative)
Re:latency v. bandwidth (Score:5, Informative)
I was just saying that what is doubled in this case is the pipe capacity and latency; the bandwidth part stays the same.
Re:Um... (Score:2, Informative)
Re:latency v. bandwidth (Score:3, Informative)
But even by the more recently accepted definition of bandwidth, you're not quite right. Latency *does* matter when we're talking about packet networking such as pigeon based transport. What if a pidgeon dies in transit? In this case it'll take you up to three hours to learn of his demise, and only then can you send the information again. So high latency + packet loss has reduced your effective bandwidth dramatically - the same happens on non-pidgeon based transports. Of course techniques such as FEC can, and are used to mitigate this. In this case I'm imaginine that pidgeon loss would be quite high, and some sort of RAIP scheme would be desirable on top of a good selective retransmission algorithm.
Also: how do you know how many pidgeons you can fit in a given amount of airspace? What if you only have ten pidgeons to work with? Here the latency is critical because you need to wait for your pidgeons to return before you can send them again with more data.
So bandwidth is not the be-all and end-all of total throughput. In many real-world situations, all the bandwidth in the world won't make your connection any work any faster.
Re:Ha! (Score:1, Informative)
4GB = 4*1024*8 = 32768Mbits
data transfer rate = 2.27Mbps = 32768/t Mbps
transfer time = t secs.
pigeon speed = x m/s = 100km / t secs. = 100000/t m/s
x : 2.27
x = 100000 / (2.27 * 32768) m/s = 1.3443867 m/s = 4.83979212 km/h
Re:Back of envalope (Score:5, Informative)
~11m/s
RFC1149 (Score:2, Informative)
From RFC1149:
[snip]
Frame Format
The IP datagram is printed, on a small scroll of paper, in hexadecimal, with each octet separated by whitestuff and blackstuff. The scroll of paper is wrapped around one leg of the avian carrier. A band of duct tape is used to secure the datagram's edges. The bandwidth is limited to the leg length.
[/snip]
See. One IP datagram, one scroll of paper. The community demands interoperability tests if CPIP is ever to become a standard
*sigh*
Google Has Used Pigeon Power for Years (Score:2, Informative)