The biggest element of this is the effect wind resistance has on cyclists. I heard Chris Boardman commenting on some cycling recently that 80% of a cyclists effort is expended on moving air. There's obviously a lot of research that goes in to this and I found an interesting paper on the subject when planning on writing this blog:
In general road races have many more than just two cyclists but I'm only going to talk about two cyclists for now. This is for two reasons:
- Firstly this occurs fairly often. At some points in road races groups of cyclists "breakaway" from the main field and have to work together to stay away.
- Secondly: the math is nicer with 2 "players".
- If both cyclists Lead then it is assumed that they basically work alone (they both suffer the effects of air drag but make the speed of the slowest cyclist).
- If both cyclist Follow then it is assumed that they don't work at all and are caught by the peleton (their escape is basically over).
- If one cyclist Leads and the other Follows then they go at the speed of the Leader and the Leader fatigues.
- Let $v_1,\; v_2$ be the speed for each cyclist.
- Let $f_1,\;f_2$ be the fatigue felt by each cyclist when suffering the effects of air drag.
We assume the following utilities:
- If both cyclists Lead then cyclist $i$ receives utility:
- If both cyclists Follow then they both receive utility:
- If one cyclist Leads and the other Follows then the Leader (w.l.o.g. let it be cyclist $1$) receives utility:
If we assume for now that $v_1=v_2=2$ and $f_1=f_2=1$ (we'll not get our hands messy with units) than our bi-matrix becomes:
Let $p$ be the probability with which cyclist 1 Leads and $q$ be the probability with which cyclist 2 Leads. It can be shown (using basic game theory and/or using this Sage interact (http://interact.sagemath.org/node/49)) that there are 3 equilibria for this game:
- $(p,q)=(1,0)$ (Cyclist 1 always Leads)
- $(p,q)=(0,1)$ (Cyclist 2 always Leads)
- $(p,q))=(.5,.5)$ (Both cyclists Lead)
In reality cyclists that break away cooperate, they talk and share the load as if they were a team. The mixed strategy above does not correspond to cooperation. It corresponds to a random strategy during which the cyclists will be randomly Leading and Following. Cooperation would be that both cyclists equally share the load (in a coordinated manner), the expected utility to both would then be 1.5 (both cyclists are better off if they cooperate - this is a notion quite close to my research, some screencasts of my talks are available here).
We'll continue to consider the situation in terms of a non-cooperative game as it gives an insight as to why cyclists cooperate.
We now assume that the cyclists are different. Perhaps they have escaped on a mountain stage and one of them is a better mountain rider. Let's take a look at the probabilities of Leading in a mixed strategy equilibria for varying $v_1$ (i.e. how fast the 1st cyclist is):