[math-fun] Tour de France/bike racing/peloton equations?
With the Tour de France now running (Monday is a rest day), watching the race brought back some mathematical thoughts that I have had over the years. I did a Google search on both regular Google & Google Scholar, but wasn't able to find anything about the following problem, although I did find some interesting discussions about the game theory of 2 cyclists sprinting against one another and/or slipstreaming behind one another. This is a classical case in cycle track racing with 2 people. I'm curious about the possible modeling the entire peloton of 100-200 riders as a kind of gas and/or liquid, moving through another (much lighter) gas, in such a way that some elements of the behavior of the entire ensemble of cyclists is captured. Anyone who has watched bike racing for more than a few moments quickly notices that most of the cyclists coalesce into blobs called "pelotons". This is because it is much more efficient (requiring perhaps only 66% as much wattage) to travel in groups, sharing the load of breaking the wind, than it is to travel individually. However, blobs occasionally break off ("break away") because they want to go faster, with larger groups being able to go faster more efficiently than smaller groups. I was thinking that there ought to be some way to model this behavior in an idealized way. One idealization might have the cyclists simply _falling_ straight down through the air, but being able to move sideways to form slipstreaming groups (parachuting cyclists??). In such a model, a more energetic cyclist simply becomes a heavier cyclist. However, he can still team up with another cyclist & fall even faster. Thus, although an individual cyclist might reach a terminal velocity V_T, two such cyclists teamed together might reach a faster terminal velocity V_T'>V_T, and an even larger ensemble of cyclists might reach a terminal velocity V_T">V_T'. There needs to be a kind of "attractive force" between cyclists which tends to clump them together, but this force is quite short-range because the ability of one cyclist to shade another from the wind does not extend that far behind him/her. So if two cyclists are pulled apart far enough (some distance D_0), then the force falls completely to zero. One could possibly model this force as a Gaussian if this made the problem simpler, because the tails of the Gaussian are so small. One could also conceive that if a sufficiently massive (i.e., powerful) cyclist were to be grouped with a much less massive (i.e., weak) cyclist, the weak cyclist might still not be powerful enough to keep up with the massive cyclists even with the advantage of being shaded from the wind. We also have a problem with the modelling of how the wind-shielding is shared among the cyclists. One could conceive that the peloton is some kind of gas, where the molecules/cyclists are bouncing around inside the peloton, but lose energy when they reach the "surface" of the blob & become exposed to the wind. They then fall slower than their associates until the "attractive force" (absence of retarding force) allows them to catch up again. This should at least ensure that the peloton assumes the correct shape, which is a very long, thin (very small multiple of D_0) line. The attractive force needs to be large enough that the peloton is, in some sense, "held back" by the less massive/less capable cyclists, and moves at a slower speeds as a result. The problem is that any larger ensemble should almost _always_ fall faster than a smaller ensemble, which would eliminate the possibility of breakaways entirely. Perhaps there are several different species of cyclist molecules, with different masses & attractive forces. (Assume for the moment that the cyclists don't form into color-coded teams!!) I.e., some cyclist molecules are "better"/"fitter", so they have slightly heavier masses (they have larger terminal velocities even when on their own) and slightly smaller attractive forces (they are more willing to break away). Then if statistically a group of these more massive, less attractive cyclists molecules found themselves together at the front of the peloton, then they could indeed "break away", and possibly fall faster than the larger peloton. --- I was hoping for an extremely simply model with only perhaps 1-3 parameters which could explain some of the simpler aspects of peloton behavior. As you can see, I haven't completely worked out the model, but I thought that since bike racing is over 100 years old, some mathematician/physicist must have looked at this type of model before. Perhaps someone on this list can provide a pointer?
While not necessarily germane to your approach, perhaps this online paper is of use: http://danpat.net/docs/brekaway.pdf
Thanks very much, Hans! This paper is quite germane, although it is trying to be an accurate model, while I'm merely trying to capture the essential behavior. --- Consider a one-dimensional system where the cyclist "molecules" are falling vertically, and can pass one another without any collisions or interactions with one another. However, the cyclist molecule at the front bears the full air resistance, while the cyclist molecules further behind are shaded to some extent. According to the paper Hans referenced, the shading is quite short-range; it falls to zero after about 3 meters. We now consider 1, 2, 3, etc. numbers of cyclist molecules. Start with a single cyclist molecule falling through air. According to Wikipedia, an object of the density of water falling through the air reaches a terminal velocity of approx. 90*sqrt(d) meters/second ~= 201*sqrt(d) mph, where d is the approx diameter of the object. Wikipedia claims the terminal velocity of a human body is approx 70 m/s ~= 157 mph, terminal velocity of a cat approx 40 m/s ~= 89 mph, the terminal velocity of a bird approx 20 m/s ~= 45 mph. http://en.wikipedia.org/wiki/Drag_%28physics%29 http://en.wikipedia.org/wiki/Free_fall Consider the differential equation, where D=rho*Cd*A, rho=density of air, Cd=drag coefficient, A=cross sectional area. m * dv/dt = 1/2 * D * v^2 - m * g Solving, we get v as a function of t: v(t) = - v_T * tanh(g*t/v_T), where v_T = sqrt(2*m*g/D) Integrating, we get y as a function of t: y(t) = y_0 - v_T^2/g * log(cosh(g*t/v_T)) Consider the following thought experiment with two identical cyclist molecules. One cyclist molecule is already falling at its terminal velocity v_T, and as it passes another cyclist molecule at t=0, the other cyclist starts accelerating at g (assume no shading). While the second cyclist will never catch up, it will asymptote falling at v_T at distance ln(2)/g*v_T^2 behind. Furthermore, this distance can be covered at speed v_T in time ln2/g*v_T; this time is a kind of "time constant": the second cyclist reaches 60% of terminal velocity within this time constant, and at 4 times this time constant, the second cyclist exceeds 99% of terminal velocity. For reference, the difference in energy between the two cyclists molecules will asymptote at ln2*m*v_T^2. If we use v_T=70m/s, then the distance will be approx 346 meters and the time constant will be approx 4.95 secs. If we use v_T=13m/s (approx. 29mph), then the distance will be approx 11.94 meters and the time constant will be approx 0.919 seconds. Of course, no real cyclist is powerful enough to accelerate that fast, so real distances and time constants would be somewhat larger. This thought experiment demonstrates a "worst-case" scenario in which a cyclist gets no shading and starts with a velocity differential of -v_T to the leading cyclist. (A true "worst-case" scenario is one in which a cyclist has to stop for 30 seconds or so to change a flat, and must then catch up to the peloton by applying substantially greater power than normal to get back to the peloton; during this time, the cyclist is unshaded, unless he/she is lucky enough to have a teammate come back to share the load.) Now if the cyclist "A" in front is travelling at terminal velocity v_T and the cyclist "B" behind is shaded, then the following cyclist B will accelerate relative to the front cyclist A and eventually pass him/her A. Since A was already travelling at terminal velocity, B will be travelling at greater than terminal velocity when B passes A, so after the pass, B will start decelerating and (due to shading) A will start accelerating. Now, since A is accelerating from v_T and B is decelerating from a velocity greater than v_T, A will now pass B again, but this time they will both be going faster than v_T. Clearly, the average speed of the two cyclists will converge at some speed slightly faster than v_T, as the time between passes of A & B converges to zero. The details of whether the "cycle" of A&B passings converge, become cyclic, or diverge, depends upon the details of the shading function and the effective "inertial mass" of the riders. If the passings diverge, then at some point the shading function falls to zero and the cyclists become independent & never pass again. If the passings converge, then the cyclists asymptotically ride side by side at some terminal velocity greater than v_T. If the passings become cyclic, then the cyclists exchange places every n seconds and achieve an average speed slightly larger than v_T. It would be a lot of fun to find the conditions under which the passings become "cyclic", so that the riders "orbit" one another (in phase space), and the behavior of sharing the lead is an emergent behavior! At 05:35 AM 7/12/2010, Hans Havermann wrote:
While not necessarily germane to your approach, perhaps this online paper is of use:
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. At 02:00 PM 7/17/2010, Henry Baker wrote:
Thanks very much, Hans! This paper is quite germane, although it is trying to be an accurate model, while I'm merely trying to capture the essential behavior.
---
Consider a one-dimensional system where the cyclist "molecules" are falling vertically, and can pass one another without any collisions or interactions with one another. However, the cyclist molecule at the front bears the full air resistance, while the cyclist molecules further behind are shaded to some extent. According to the paper Hans referenced, the shading is quite short-range; it falls to zero after about 3 meters.
We now consider 1, 2, 3, etc. numbers of cyclist molecules.
Start with a single cyclist molecule falling through air.
According to Wikipedia, an object of the density of water falling through the air reaches a terminal velocity of approx. 90*sqrt(d) meters/second ~= 201*sqrt(d) mph, where d is the approx diameter of the object. Wikipedia claims the terminal velocity of a human body is approx 70 m/s ~= 157 mph, terminal velocity of a cat approx 40 m/s ~= 89 mph, the terminal velocity of a bird approx 20 m/s ~= 45 mph.
http://en.wikipedia.org/wiki/Drag_%28physics%29
http://en.wikipedia.org/wiki/Free_fall
Consider the differential equation, where D=rho*Cd*A, rho=density of air, Cd=drag coefficient, A=cross sectional area.
m * dv/dt = 1/2 * D * v^2 - m * g
Solving, we get v as a function of t:
v(t) = - v_T * tanh(g*t/v_T), where v_T = sqrt(2*m*g/D)
Integrating, we get y as a function of t:
y(t) = y_0 - v_T^2/g * log(cosh(g*t/v_T))
Consider the following thought experiment with two identical cyclist molecules. One cyclist molecule is already falling at its terminal velocity v_T, and as it passes another cyclist molecule at t=0, the other cyclist starts accelerating at g (assume no shading). While the second cyclist will never catch up, it will asymptote falling at v_T at distance ln(2)/g*v_T^2 behind. Furthermore, this distance can be covered at speed v_T in time ln2/g*v_T; this time is a kind of "time constant": the second cyclist reaches 60% of terminal velocity within this time constant, and at 4 times this time constant, the second cyclist exceeds 99% of terminal velocity. For reference, the difference in energy between the two cyclists molecules will asymptote at ln2*m*v_T^2.
If we use v_T=70m/s, then the distance will be approx 346 meters and the time constant will be approx 4.95 secs. If we use v_T=13m/s (approx. 29mph), then the distance will be approx 11.94 meters and the time constant will be approx 0.919 seconds. Of course, no real cyclist is powerful enough to accelerate that fast, so real distances and time constants would be somewhat larger.
This thought experiment demonstrates a "worst-case" scenario in which a cyclist gets no shading and starts with a velocity differential of -v_T to the leading cyclist. (A true "worst-case" scenario is one in which a cyclist has to stop for 30 seconds or so to change a flat, and must then catch up to the peloton by applying substantially greater power than normal to get back to the peloton; during this time, the cyclist is unshaded, unless he/she is lucky enough to have a teammate come back to share the load.)
Now if the cyclist "A" in front is travelling at terminal velocity v_T and the cyclist "B" behind is shaded, then the following cyclist B will accelerate relative to the front cyclist A and eventually pass him/her A. Since A was already travelling at terminal velocity, B will be travelling at greater than terminal velocity when B passes A, so after the pass, B will start decelerating and (due to shading) A will start accelerating. Now, since A is accelerating from v_T and B is decelerating from a velocity greater than v_T, A will now pass B again, but this time they will both be going faster than v_T. Clearly, the average speed of the two cyclists will converge at some speed slightly faster than v_T, as the time between passes of A & B converges to zero.
The details of whether the "cycle" of A&B passings converge, become cyclic, or diverge, depends upon the details of the shading function and the effective "inertial mass" of the riders. If the passings diverge, then at some point the shading function falls to zero and the cyclists become independent & never pass again. If the passings converge, then the cyclists asymptotically ride side by side at some terminal velocity greater than v_T. If the passings become cyclic, then the cyclists exchange places every n seconds and achieve an average speed slightly larger than v_T.
It would be a lot of fun to find the conditions under which the passings become "cyclic", so that the riders "orbit" one another (in phase space), and the behavior of sharing the lead is an emergent behavior!
At 05:35 AM 7/12/2010, Hans Havermann wrote:
While not necessarily germane to your approach, perhaps this online paper is of use:
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.
The power (or "mass", in your falling-through-air analogy) of the cyclists changes *over* *time* and from one cyclist to another due to their strategies. As in running and skating, etc., competitors apply a strategy to budget their energy over time. On-the-spot changes in strategy and psychological effects are usually involved (the tortoise and the hare). These energy strategies are a large part of the cause of breakaways. Without modeling them, your model with not exhibit the typical behaviour of a peloton (even ignoring breakaways) when it stretches and occasionally fissions or gives birth to small groups. If you force the fissioning behaviour through changes to the simple model, you'll have a peloton that fissions for no relevant (strategic) reason, and therefore will split too often once an energy budget strategy model is added. On Mon, Jul 12, 2010 at 00:36, Henry Baker <hbaker1@pipeline.com> wrote:
[...] Anyone who has watched bike racing for more than a few moments quickly notices that most of the cyclists coalesce into blobs called "pelotons". This is because it is much more efficient (requiring perhaps only 66% as much wattage) to travel in groups, sharing the load of breaking the wind, than it is to travel individually. However, blobs occasionally break off ("break away") because they want to go faster, with larger groups being able to go faster more efficiently than smaller groups. [...] The problem is that any larger ensemble should almost _always_ fall faster than a smaller ensemble, which would eliminate the possibility of breakaways entirely.
Perhaps there are several different species of cyclist molecules, with different masses & attractive forces. (Assume for the moment that the cyclists don't form into color-coded teams!!) I.e., some cyclist molecules are "better"/"fitter", so they have slightly heavier masses (they have larger terminal velocities even when on their own) and slightly smaller attractive forces (they are more willing to break away). Then if statistically a group of these more massive, less attractive cyclists molecules found themselves together at the front of the peloton, then they could indeed "break away", and possibly fall faster than the larger peloton. [...]
-- Robert Munafo -- mrob.com
participants (4)
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Fred lunnon -
Hans Havermann -
Henry Baker -
Robert Munafo