- Details
- Written by Ivey Brent Laminack Ivey Brent Laminack
- Category: Technology Technology
- Published: 29 November -0001 29 November -0001

### First the Megabytes

Our minivan is a 1988 Plymouth Voyager (not the Grand model). Removing the back two seats gives a cargo space of 48 inches wide by 42 inches high by 72 inches long. CDROMs seem to fit nicely in a 5" X 5" footprint. At MicroCenter they were selling spindles of 100 CDROMs in a case about 8" high. Rounding this says that on the floor of the minivan would fit a matrix of 9 by 14 CDROMs and we could stack them 600 high if we had a suitable spindle. So our minivan could hold 9X14x600=75,600 CDROMs. What is the data capacity of such? At 650 Megabytes per CDROM, we come out with 75,600X650=49,140,000 Megabytes, or 49,140 Gigabytes or 49.14 Terabytes.

### Now the Seconds

Consider a one-way trip from Washington, DC to San Francisco California (MAE East to MAE west). My road atlas gives 2890 miles (about 21 miles shorter than New York City to Los Angeles). Assuming we have two drivers and haul as fast as legally possible (averaging 50 MPH) this gives us a driving time of 2890/50 = 57.8 hours. This is reasonable, because the winning time in the Fireball Baker Sea-to-Shining-Sea Memorial race is typically under 36 hours.(This is the race that has inspired such cinematic bombs as Gumball Rally and The Cannonball Run) How many seconds is this? 57.8 hours X 60 minutes/hour X 60 seconds/minutes = 208,080 seconds.

### Bandwidth = Megabytes / Second

49,140,000 Megabytes / 208,080 seconds = 235 Megabytes/Second.How does this compare with fiber?This illustrates that bandwidth isn't everything, one needs low latency as well to be useful. A week of turn-around time of data going from coast to coast and back isn't very useful, no matter what its bandwidth.