3 ms·
That's not how the math works out, it isn't linear. The distance it takes to stop anything is dominated by a term quadratic in the velocity. Thus the average sa
by BlackFly 5y ago
That's not how the math works out, it isn't linear. The distance it takes to stop anything is dominated by a term quadratic in the velocity. Thus the average safe traffic density decreases with the velocity like v^-2. Since the throughput will just be the velocity times the traffic density, throughput roughly decreases proportional to 1/v. (In the zero velocity limit, physical size terms would dominate, so you don't get infinite throughput.)
https://www.omnicalculator.com/physics/stopping-distance https://www.omnicalculator.com/physics/stopping-distance
So if you do the math, bikes take up less time-space, or said better: achieve higher throughput. They are multiple times more dense at their travelling speeds. The only real way out of this would be to travel at unsafe distances.
- sandworm101 5y agoYou assume single passenger vehicles. Do the math again for a 50-person busses. Compared to that, bikes and cars are relatively similar. Even four-person cars, with say 4 seconds between them (one person per second) better single-passenger bikes traveling in pairs. Bikes rarely ever maintian such separations, genrally leaving 7+ seconds between pairs when at speed due to bikes having reduced stopping times from top spèed.
- jka 5y agoIt may be a reasonable assumption; what are typical vehicle occupancy rates? Also, a bicycle-majority traffic mix wouldn't have to conflict with the presence of high-passenger-capacity buses. That may be true that cars can have shorter time-distance separation, but I think there's a gap in the modeling there with regard to group travel: If you have a four-passenger car and compare that to the same group of people traveling together by bicycle, the time-distance separation factor disappears.