Of course, if they eventually ride side-by-side, there is no shading and they slow down ... This comment is not quite as trite as it may appear, since rather than riding in a line, the peloton does in fact assume an extended teardrop shape. I wonder if --- at the leading edge anyway --- there is an effect similar to the wave-coupling which leads to geese flying in a V-formation? WFL On 7/20/10, Henry Baker <hbaker1@pipeline.com> wrote:
A more accurate 2-person cycling simulation than my "falling cyclist" is given by a "constant power" approximation, so the differential equation is now
m*dv/dt = P/v - (1/2)*D*v^2
where the terminal velocity is v_T=(2*P/D)^(1/3).
Semi-reasonable numbers are m=75kg, v_T=9m/s, P=216watts, D=0.5926.
While the above differential equation appears not to be solvable in any reasonable closed form, we can approximate solutions of small deviations from v_T. When this is done, we find that these small deviations decay exponentially to 0 with a time constant
t_C = m*v_T^2/(3*P) ~ 9.375 seconds.
Associated with this time constant is a distance constant
t_C*v_T ~ 84.375 meters.
Also, an energy constant
t_C*P ~ 2.025 kJ.
--
Suppose the shaded cyclist has a drag D=0.432 instead of D=0.5926. Then the terminal velocity for the shaded cyclist will be v_T=10m/s instead of 9m/s. The time constant for the shaded cyclist is 11.574 secs and the distance constant for the shaded cyclist is 115.74 meters.
We can now estimate a speed for the two cyclists "working together" as a mini-peloton. (Actually, they aren't "working together" at all, because their behavior emerges from the equations.)
If the leading cyclist is going faster than his/her terminal velocity (9m/s), then he/she will slow down with an exponential decay towards 9m/s. If the following cyclist is going slower than his/her terminal velocity (10m/s), then he/she will speed up with an exponential decay towards 10m/s. The average velocity of the two cyclists should end up somewhere in the range of 9-10m/s.
Since there are only 2 cyclists, and they are both identical, symmetry requires that both spend 50% of the time in the lead and 50% of the time being shaded. But this is equivalent to facing a drag constant which is an arithmetic _average_ of the two drag constants, hence D~0.5123. So we can now calculate the average terminal velocity as (2*216/0.5123)^(1/3)=9.448m/s. Thus, the velocity of this mini-peloton is _not_ the arithmetic average of the two terminal velocities, but slightly less than that.
In this simple model, the "dwell time" of each cyclist in the lead gets shorter & shorter, so the cyclists exchange places faster & faster until they are riding "together" -- the distance has converged to zero. Notice that there is no "coordination" between the two cyclists -- they simply produce power at a constant unvarying wattage.