W4.3 Critical mass

Last year while I was visiting Vancouver, I took part in a cycle ride which takes place on the last Friday evening of every month, entitled Critical Mass. Several thousand riders took part and the culmination of the ride was the pack of cyclists filling all lanes on the Lions Gate bridge (Vancouver’s version of the Golden Gate Bridge), dismounting and raising their bikes above their heads.

Police patrol the event on bikes. I was talking to an officer while on the bridge and (even managing to think about things like this while on holiday) I asked him if we should be worried about the load the cyclists were placing on the bridge. My gut feeling was that this was possibly more than that for the more conventional use of the bridge, which was carrying vehicles.

The police officer looked at me as if I was mad. But was my intuition right or not?

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Hint

The key concept here is to consider the mass density of vehicles using the bridge (ie mass per unit area)

Consider limiting cases for cars of a gridlocked traffic jam (the highest density of vehicles) and freely moving traffic (lowest) and compare the mass density for each of these with stationary bicycles and riders.

 

Solution

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Solution

We are going to need to make some estimates (but that shouldn’t be a surprise to you by now)....

Cars on the bridge; gridlock

  • The surface area taken up by a car is about 2.5×4m=10m2
  • An average car weighs about 2000kg
  • In a gridlock situation cars will be nose to tail in each lane, so a typical area occupied will be about 5×5m=25m2 (note that although there’s not much space between cars in the same lane, there’s usually space in between lanes .....)
  • So in this case the mass density, let’s call it ρ is given by
    $$\rho_{jam} = \frac{2000}{25} = 80 kg m^{-2}$$

Cars on the bridge; free-flowing traffic

  • In free flowing traffic cars will be spaced by at least 10m if travelling at a reasonable speed. (We could calculate the recommended separation according to the Highway Code, but that is probably overkill here....)
  • So the typical area occupied will be about 5×15m=75m2
  • So in this case the mass density is given by
    $$\rho_{free} = \frac{2000}{75} \approx 27 kg m^{-2}$$

Bikes on the bridge; all riders stopped

  • The surface area taken up by a bike is a long thin rectangle - about 1×2m=2m2
  • An average bike rider and bike weighs about 100kg
  • So in this case the mass density is given by
    $$\rho_{bike} = \frac{100}{2} = 50 kg m^{-2}$$

Now, the bikes will probably not pack that close together, but even if they were packed at 50% density, that still puts the load on the bridge comparable to that when full of free flowing traffic! Physics 1, Police, nil.

Good job it was on bikes, really. If we had been pedestrians on the bridge, we could have packed far tighter and easily exceeded the load on the bridge at total gridlock.......

And just in case you thought it was all made up..... this is what it looked like
Graphic - Critical Mass photo