My model was 1-dimensional with the riders being 0-dimension points, so strictly speaking, there's no such thing as "side-by-side", and the two riders exchange places infinitely often, closer & closer. In the real (2-dimensional) world of cycling, the peloton assumes an extended teardrop shape at the rear, but for some reason, it does _not_ have a teardrop shape at the front. There's typically a line of 3-5 riders out in front -- what you might expect for a supersonic peloton(!). I'm having trouble imagining a model in which this shape is optimal, but I'll take a stab at it. I think when the peloton assumes this shape, the 3-5 riders in the line at the front are typically always from the same team, and have some strategic reason for trying to speed up the peloton as a whole. The rest of the peloton is happy to have them assume the lead, but the rest of the peloton doesn't want to do as much work. So the team at the front is -- in effect -- doing a team time trial, in which the 3-5 riders stay in the front and cycle through their 3-5 riders in succession, but no one else participates in this team time trial. So one side of the road must remain clear to allow the lead rider to drift backwards & reassume the 3rd, 4th or 5th position. Since the road has an equal chance of turning to the right or to the left, both sides remain clear to give the lead rider the opportunity to drift backwards on either side. Those not on the lead team time trial are happy to participate in whatever shade they can get from that lead team, so those riders form a V-shape go ing backwards as the front shape of the peloton. I know nothing about the V-formation of migrating birds, but it might be some similar effect, which requires little or no planning/strategy/thinking on the part of the birds. If the following birds get sucked into vortices from the birds further up the line, then it might actually require _extra_ effort to get out of the vortex. At 07:42 PM 7/19/2010, Fred lunnon wrote:

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.