Pigeons vs The Internet
When you are trying to transfer a lot of data to another computer there are sometimes quicker ways than just sending it over the internet. Sending hard drives by post can be a valid method for large amounts of data. I just spent a lunchtime crunching the numbers for how much data you would have to send from Hereford to Paris for carrier pigeons to be a faster method of data transfer than the internet. These locations were originally my place of work and New York but I noticed that pigeons would have trouble with a multiday flight over the Atlantic so Paris it became.
The logic is that data storage has become so light and compact that even if it is moved slowly the pigeons might stand a chance. Because they are slow then for small amounts of data it will always be quicker to use the Internet, so the question becomes: how much data would you need to be sending from Hereford to Paris for it to be worth sending a pigeon?
Let's run some numbers. A pigeon can hold up to 75 grams with training. (I'm quoting from a website. What this training entails I'm not entirely sure.) After a brief look around Micro SD cards seem to be around 0.5 grams and can hold up to 256 GB. Therefore a pigeon can carry 150 SD cards or about 38.4 TB (terrabytes).
To compare this with the capacity of an Internet transfer then we need the time for the birds to fly there. A racing pigeon can maintain a 92.5 MPH average over a 400 mile race. Paris is just under this limit, so it seems a fair estimation and it results in a time of 3.4 hours. In seconds this works out as 12,240.
However this doesn't give the whole picture because we still have to load the SD cards up with the data and then unload it at the other end. SD card upload and download times are both around 95 MB/S (Megabytes per Second). So the total time to get the data (D megabytes) across would be D(1/95+1/95)+12240.=2D/95+12240.
For Internet transfers Hereford gets upload speeds of about 1.25 MB/S and Paris gets a Download speed of about 10 MB/S. Since the data can be downloaded as soon as it goes up that is the bottleneck. The ping of about 0.3 seconds is also irrelevant as it is tiny in comparison. So really the time to transfer D data (for D of a non trivial size) would be D/1.25.
Putting these two expressions together by equating the times gives 2D/95+12240=D/1.25 which solves to give D=15714 MB or about 16 GB. This is a tiny amount which is well within the capacity of a Pigeon that could hold 2000 times that.
So since pigeons are better than we thought, let's change the problem. If we wanted to transfer a whole pigeon’s worth of data (38.4 TB) then at what distance would a data transfer going close to the speed of light be faster? Adapting our equations we have D = 38.4 TB = 38400 GB. The times from both sides are 2*38400/95 + 3600*X/92.5 = 38400/1.25 where X is the distance at which both methods of transfer are about the same speed. This works out as X = 3843, or about 4000 Miles. To put this in context I'm saying that a racing pigeon and an internet transfer are about the same speed if you are trying to get close to 40 TB of data transfered then this is a map of the locus: