On modelling bicycle power for velodromes: Part I

Formulation for individual pursuits

Michael A. Slawinski∗, Raphael A. Slawinski†, Theodore Stanoev‡

September 3, 2020

Abstract

For a moving bicycle, the power can be modelled as a response to the propulsion of the centreof mass of the bicycle-cyclist system. On a velodrome, an accurate modelling of power requiresa distinction between the trajectory of the wheels and the trajectory of the centre of mass. Weformulate and examine an individual-pursuit model that takes into account the aforementioneddistinction. In doing so, we provide details of the invoked physical principles and mathematicalderivations, with an emphasis on their limitations. We assume that a velodrome consists oftwo parallel straights and two semicircular arcs. We neglect the effects of the track inclinationalong the straights and assume the track inclination along the curves to be constant. For eithersegment, we consider two distinct black-line speeds. For the latter, the lean-angle expressionis derived based on a noninertial frame of the cyclist. Among conclusions quantified by thismodel is the fact that a constant-cadence approach to an individual pursuit does not minimizethe required power.

1 Introduction

In this article, we consider a mathematical model to account for power expenditure on velodromes.This work is a mathematization of certain aspects of studies presented by Martin et al (1998) andUnderwood (2012). The discussed model is pertinent to an individual pursuit of a cyclist followingthe black line in a constant aerodynamic position. The circumference along the inner edge of thisfive-centimetre-wide line — also known as the measurement line and the datum line — correspondsto the official length of the track.

Using this model, for a given cyclist, we can calculate the power required to achieve a desiredtime or — since the relation between power and speed is one-to-one — the time achievable with aparticular power. Also, for repeated laps, we can estimate model parameters from the power andspeed measurements. The method proposed to estimate these parameters is specific to velodromes,and is different from the circuit study presented by Chung (2012).

We begin this article by presenting a model and justifying its mathematical formulation. We illus-trate both forward and inverse applications of the model. We conclude by a discussion of results.This article contains also three appendices.

∗Memorial University of Newfoundland, Canada, mslawins@mac.com†Mount Royal University, Canada, rslawinski@mtroyal.ca‡Memorial University of Newfoundland, Canada, theodore.stanoev@gmail.com

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2 Formulation

A mathematical model to account for the power required to propel a bicycle — along a straightcourse — with speed V→ is (e.g., Danek et al, 2020a)

P = F← V→ (1)

=

gravity︷ ︸︸ ︷mg sin Θ +

change of speed︷ ︸︸ ︷ma +

rolling resistance︷ ︸︸ ︷Crr mg cos Θ︸ ︷︷ ︸

normal force

+

air resistance︷ ︸︸ ︷12 ηCdA ρ (V→ + w←︸ ︷︷ ︸

air flow speed

)2

1− λ︸ ︷︷ ︸drivetrain efficiency

V→ ,

where F← stands for the forces opposing the motion and V→ for the ground speed. In particular, mis the mass of the cyclist and the bicycle, g is the acceleration due to gravity, Θ is the slope of a hill,a is the change of speed, Crr is the rolling-resistance coefficient, CdA is the air-resistance coefficient,ρ is the air density, w← is the wind component opposing the motion, λ is the drivetrain-resistancecoefficient, η is a quantity that ensures the proper sign for the tailwind effect, w← < −V→ ⇐⇒ η =−1 , otherwise, η = 1 .

To consider a steady ride, a = 0 , on a flat course, Θ = 0 , in windless conditions, w = 0 , we write

P =Crrmg + 1

2 CdA ρ V 2→

1− λ︸ ︷︷ ︸F←

V→ . (2)

In modelling the power along a straight course, there is no distinction between the ground speed ofthe centre of mass and of any other point of the bicycle-cyclist system. The distinction appears ifthe cyclist deviates from a straight course by leaning, which becomes pronounced on a velodrome.

Let us consider a velodrome, whose black-line distance is S , and the banking angle and radius areθ and r , respectively. Also, let us assume that the centre of mass of the bicycle-cyclist system is habove the ground, without lean.1 If so, the radius of the centre-of-mass trajectory, rCoM , is shorterthan r by h sinϑ , where ϑ is the angle, measured from the vertical, at which the cyclist leans. Hence,the distance traveled — in one lap — by the centre of mass is shorter than the black line by

2π r − 2π

rCoM︷ ︸︸ ︷(r − h sinϑ) = 2πh sinϑ . (3)

Thus — neglecting a progressive leaning and straightening between the straights and the circulararcs — the distance travelled by the centre of mass, along the straights and along the curves, for asingle lap, is

S − 2πr and 2π(r − h sinϑ) ≡ 2π rCoM , (4)

respectively.

Consider a laptime,

t =S − 2πr

v→+

2πr

vx, (5)

1We assume that the position of a cyclist on a bicycle remains the same, which is a reasonable assumption for anindividual pursuit, after the initial acceleration. Hence, h is constant, and the change of the height of the centre ofmass is due only to the lean angle, ϑ .

2

Figure 1: Christiaan Huygens confident of a centripetal force, which he mathematized in 1659

where v→ and vx are the black-line speed along the straights and the black-line speed along thecurves, respectively; denotes the arithmetic average. Also, the time along the curves is

tx =2πr

vx=

2π rCoM

Vx, (6)

where Vx is the average centre-of-mass speed along the curves; v→ ≡ V→ , since — along thestraights — there is no distinction between the speed of the centre of mass and the speed of anypoint of the bicycle-cyclist system.

The lean angle of a cyclist — as illustrated in Figures 1 and 2, and entailed by expression (20), below,as well as discussed in Appendix A — is

ϑ = arctanFcpFg

, (7)

where the magnitude of the centripetal force is

Fcp =mV 2

xrCoM

, (8)

and of the force of gravity is Fg = mg ; hence,

Vx =√g rCoM tanϑ . (9)

Inserting expression (9) into expression (6), we obtain

vx = r

√g tanϑ

rCoM

. (10)

Combining expressions (5), (6) and (9), we write

t =S − 2πr

V→+ 2π

√rCoM

g tanϑ. (11)

3

Given v→ and vx , with r , S and h assumed to be known, we use equations (5) and (10) to find tand ϑ , respectively. Then, we use expression (9) to find Vx .

Following expressions (4), the proportion of distance travelled, per lap, by the centre of mass, is

1− 2πr

Sand

2π(r − h sinϑ)

S,

along the straights and the curves, respectively. Hence, according to the harmonic average, discussedin Appendix B, the average centre-of-mass speed, per lap, is

〈V 〉 =1

1

V→

1− 2πr

S

1− 2πh sinϑ

S

+1

Vx

2π(r − h sinϑ)

S

1− 2πh sinϑ

S

(12a)

=V→Vx (S − 2πh sinϑ)

S Vx + 2π(r(V→ − Vx

)− V→ h sinϑ

) . (12b)

The average power per lap is

P =1

1− λV→Vx (S − 2πh sinϑ)

S Vx + 2π(r(V→ − Vx

)− V→ h sinϑ

)︸ ︷︷ ︸〈V 〉

(13a)

Crrmg

(1− 2πr

S

)(13b)

+

Crrmg (sin θ tanϑ+ cos θ)︸ ︷︷ ︸N

cos θ + Csr

∣∣∣∣∣mgsin(θ − ϑ)

cosϑ︸ ︷︷ ︸Ff

∣∣∣∣∣ sin θ 2π

rCoM︷ ︸︸ ︷(r − h sinϑ)

S(13c)

+ 12 CdA ρ

(1− 2πr

S

)V

2

→ +2π

rCoM︷ ︸︸ ︷(r − h sinϑ)

SV

2

x

. (13d)

Herein, Csr is the coefficient of the lateral friction and∣∣ ∣∣ stands for the magnitude. If S → ∞ ,

expression (13) reduces to expression (2), as expected.

The second fraction in factor (13a) is the average centre-of-mass speed, per lap, which combines,proportionally, V→ , along the straights, and Vx , along the banks. Summand (13b) is the rollingresistance along the straights.2 Summand (13d) is the air resistance, which is a function of the

2A modern velodrome, such as Velodrome de Bordeaux-Lac, has S = 250 m , r = 23 m , θ = π/4 rad ≡ 45◦ , alongthe banks, Θ = 0.23 rad ≡ 13◦ , along the straights; the latter angle is not zero to avoid an excessive change of trackinclination and to allow for a gentle slope transition between the banks and the straights. We could refine our modelby including (

Crr cos2Θ + Csr sin2Θ)mg

(1−

2πr

S

),

to account for the force of lateral friction along the straights, which would be analogous to summand (13c), with

4

Figure 2: Force diagram: Inertial frame

proportions between V2

→ , along the straights, and V2

x , along the banks.3

To formulate summand (13c), we use the relations among the magnitudes of vectors N , Fg , Fcpand Ff , illustrated in Figure 2. In accordance with Newton’s second law, for a cyclist to maintaina horizontal trajectory, the resultant of all vertical forces must be zero,∑

Fy = 0 = N cos θ + Ff sin θ − Fg . (14)

In other words, Fg must be balanced by the sum of the vertical components of normal force, N ,and the friction force, Ff , which is parallel to the velodrome surface and perpendicular to theinstantaneous velocity. Depending on the centre-of-mass speed and the radius of curvature for thecentre-of-mass trajectory, if ϑ < θ , Ff points upwards, in Figure 2, which corresponds to its pointingoutwards, on the velodrome; if ϑ > θ , it points downwards and inwards. If ϑ = θ , Ff = 0 . Sincewe assume no lateral motion, Ff accounts for the force that prevents it. Heuristically, it can beconceptualized as the force exerted in a lateral deformation of the tires.

For a cyclist to follow the curved bank, the resultant of the horizontal forces,∑Fx = −N sin θ + Ff cos θ = −Fcp , (15)

is the centripetal force, Fcp , whose direction is perpendicular to the direction of motion and pointstowards the centre of the radius of curvature. According to the rotational equilibrium about thecentre of mass, ∑

τz = 0 = Ff h cos (θ − ϑ)−N h sin(θ − ϑ) , (16)

where τz is the torque about the axis parallel to the instantaneous velocity, which implies

Ff = N tan(θ − ϑ) . (17)

ϑ = 0 . However, since Θ is small, cos Θ ≈ 1 and sin Θ ≈ 0 , we choose to consider only

Crrmg

(1−

2πr

S

),

which is summand (13b).3We could refine the model by including the effect of air resistance of rotating wheels (Danek et al, 2020a, Ap-

pendix D), which would require introducing another resistance coefficient to summand (13d), if the wheels are thesame, or two coefficients, if they are different.

5

Substituting expression (17) in expression (14), we obtain

N =mg

cos θ − tan(θ − ϑ) sin θ= mg (sin θ tanϑ+ cos θ) . (18)

Using this result in expression (17), we obtain

Ff = mg (sin θ tanϑ+ cos θ) tan(θ − ϑ) = mgsin(θ − ϑ)

cosϑ. (19)

Since the lateral friction, Ff , is a dissipative force, it does negative work. Hence, the work doneagainst it — as well as the power — needs to be positive. For this reason, in expression (13c), weconsider the magnitude of Ff .

To relate Fcp and ϑ , we use results (18) and (19) in expression (15), to obtain

Fcp = N sin θ − Ff cos θ = mg tanϑ , (20)

which is tantamount to expression (7). Examining expressions (8) and (20), we see that the leanangle is a function of the centre-of-mass speed and of the radius of curvature for the centre-of-masstrajectory; it is independent of mass or the track inclination.

Figure 3: Fg in terms of vertical components of N and Ff

In terms of solutions (18) and (19), expression (14) — as a function of ϑ , for a fixed value of θ— isshown in Figure 3. Fg is constant, as required. Also, as required, Ff = 0 and N cos θ = Fg , atϑ = θ . For ϑ < θ , Ff points outwards, hence — in accordance with Figure 2 — it is positive. Forϑ > θ , Ff points inwards and, hence, is negative. The crossing of two curves corresponds to ϑ atwhich the vertical components of N and Ff are equal to one another.

3 Numerical example

3.1 Forward model

4Let us consider the following values. For the bicycle-cyclist system, m = 111 kg , h = 1.2 m ,CdA = 0.2 m2 , Crr = 0.002 , Csr = 0.003 and λ = 0.02 . For the velodrome, S = 250 m , r = 23 mand θ = 0.7505 rad ≡ 43◦ . For the external conditions, g = 9.81 m/s2 and ρ = 1.225 kg/m3 .

4For consistency with power meters, whose measurements are expressed in watts, W , which are kg m2/s3 , we usethe SI units for all quantities. Mass is given in kilograms, kg , length in meters, m , and time in seconds, s ; hence,area is in m2 , speed is in m/s acceleration in m/s2 and force in kg m/s2 =: N ; angles are in radians, rad , as well asin degrees,◦ .

6

Let the laptime be such that, according to expression (5), the corresponding black-line speed isv→ = vx = V→ = 12 m/s ; a constant black-line speed is tantamount to a constant cadence, which isa common approach for a workout or even for a pursuit. In accordance with expressions (6) and (9),ϑ = 0.555468 rad ≡ 31.8260◦ , and Vx = 11.6698 m/s . Hence, in accordance with expression (13),P = 229.6723 W . Also, in accordance with expression (12), the average centre-of-mass speed, perlap, is Vx < 〈V 〉 = 11.8091 m/s < V→ , as expected.

The values of summands (13b), (13c) and (13d), which represent the forces opposing the movement,are, respectively, 0.9189 N , 1.3195 N and 16.8213 N . The air resistance, which is the third summand,has the dominant effect, with respect to the other summands, which are associated with the effectof the wheel contact with the velodrome surface.

If the centre of mass is not taken into account, expression (2) results in P = 242.6672 W , which isan overestimate. The values of the first and second summands in the numerator of expression (2)are 2.1778 N and 17.6400 N , respectively. Therein, the first summand corresponds to the effect ofthe wheel contact with the surface and the second to the effect of the air resistance.

The similarity between the values obtained with expressions (2) and (13) is a supportive evidence forthe correctness of refinements provided by the latter and an indication of a reasonable accuratenessof the former, in spite of its simplicity. However, recognizing the difference between these values isconsistent with the attention to marginal gains that underpin the Team GB dominance in velodromecompetitions (e.g., Slater, 2012).

For this numerical example, N = [−857.4810 , 919.5203] , Fg = [0 ,−1088.91] Fcp = [−675.8359 , 0]and Ff = [181.6451 , 169.3897] . These forces are illustrated in Figure 2; the positive directions areupward and rightward. As required, the resultant of the vertical forces is zero, and the resultantof the horizontal forces is equivalent to Fcp . Also, the orientation of Ff is θ = 0.7505 rad ≡ 43◦ ,which means that it is parallel to the velodrome surface, as required.

If we consider the same laptime, namely, t = 250/12 = 20.8(3) s , and, using expression (5), wearbitrarily choose v→ = 11.7364 m/s ≡ V→ and vx = 12.2 m/s , which is consistent with empiricalexaminations of the difference between the black-line speed along the straights and along the curves,we obtain ϑ = 0.5701 rad ≡ 32.6643◦ and Vx = 11.8565 m/s , which results in P = 229.2440 W <229.6723 W . This result is consistent with the fact that the same average power results in a higherblack-line speed along the curves than along the straights due to the shorter distance covered by thecentre of mass. It also implies that — to achieve a given laptime — maintaining a constant black-linespeed, which is tantamount to a constant cadence, does not minimize the required average power.Nor does it imply that the minimum average power results from maintaining a constant instantaneouspower. In other words, a minimization of the average power to achieve a given laptime might notbe tantamount to maintaining either a constant black-line speed or a constant power.

3.2 Inverse problem

For this numerical example, expression (13) can be written as

P = 229.6723 W = 993.2258CdA

1− λ︸ ︷︷ ︸X

+11530.0934Crr

1− λ︸ ︷︷ ︸Y

+1124.4934Csr

1− λ︸ ︷︷ ︸Z

. (21)

In contrast to expression (1), for expression (2), and for its extension, given by expression (13),the resistance coefficients appear only as ratios. Hence, even with many independent equations, wecannot obtain — as an inverse solution — the values of CdA , Crr , Csr and λ , but only the ratios,X , Y and Z .

7

To obtain the values of X , Y and Z, we perform the least-squares fit of ten equations analogous toequation (21), with V→ ∈ (11.5, 12.5) , whose matrix representation is

250.2814221.2244246.4302207.1293210.2785234.5504231.1933225.5622221.3932235.9688

=

1090.8398 12040.9534 1008.6335953.2952 11315.5242 1169.92591072.5787 11946.7665 1030.7921886.7918 10950.0690 1242.7456901.6366 11032.5717 1226.81691016.3057 11652.5725 1097.69211000.4207 11568.3931 1116.1798973.7925 11426.0999 1146.7564954.0928 11319.8441 1169.03051023.0198 11687.9987 1089.8239

XYZ

. (22)

The least-squares solution of system (22) is X = 0.204082 , Y = 0.002041 , Z = 0.003061 . Sinceλ = 0.02 , we obtain CdA = 0.2 m2 , Crr = 0.002 , Csr = 0.003 , as expected.

For the measured values of P — as opposed to the modelled ones, for which three equations sufficeto solve for X , Y and Z — a redundancy of the laptime information allows us to estimate them,and to obtain statistical information about the empirical adequacy of a model, which, howeversophisticated, remains only a mathematical analogy for a physical realm. This redundancy couldcorrespond to different laptimes during a single ride.

An insight into the consistency of information can be gained by writing each equation of system (22)as

1 =a

P

CdA

1− λ︸ ︷︷ ︸X

+b

P

Crr

1− λ︸ ︷︷ ︸Y

+c

P

Csr

1− λ︸ ︷︷ ︸Z

,

and plotting a/P , b/P and c/P . For system (22), they are collinear. For measurements, thedeparture from the collinearity is indicative of the quality of the model and of the measurement errors.Within a model, such a plot can be used to study the sensitivity of X , Y and Z to perturbations.

To estimate CdA , Crr and Csr , the value of λ needs to be given independently or be assumed. It iscommonly accepted that, for high-quality track bicycles, λ ∈ (0.01, 0.03) . Also, if the power meteris in the rear hub, as opposed to being in the pedals or the bottom bracket, λ ≈ 0 , since the effectof the resistance of the drivetrain — which includes the chainring, chain and sprocket — upon themeasuring device is nearly eliminated (e.g., Chung, 2012).

In general, given the laptime — and assuming v→ to be known — we can use equation (11) to find ϑ ,which in turn can be substituted into equation (9) to find Vx , which can then be substituted intoequation (6) to find vx . In other words, given the laptime and the speed along the straights, thelean angle is implicitly determined, and in turn determines the centre-of-mass and black-line speedsalong the curves.

4 Conclusions

Expression (13), together with expressions (5), (6) and (9), allows us to calculate the power requiredto achieve a desired individual-pursuit time or the time achievable with a particular power. Italso allows us to quantify the effects of the bicycle-cyclist weight, air resistance, rolling resistance,drivetrain resistance and lateral friction, as well as of the velodrome size, steepness of its banksand tightness of its curves. Furthermore, a quantification of these effects lends itself to a study ofoptimization of a cyclist’s effort (Danek et al, 2020b, Appendix A).

8

Our formulation and examination of an inverse solution for expression (13) shows that we can inferonly the ratios, CdA/(1 − λ) , Crr/(1 − λ) and Csr/(1 − λ) . Nevertheless, the proposed solutionallows us to gain an insight into the consistency between the measurements and the model.

As illustrated by the presented results, accurate inferences based on the power-meter measurementson a velodrome require a distinction between the trajectory of wheels, which we assume to coincidewith the black line, and the trajectory of the centre of mass. The forces involved and relationsamong them, as well as other entailed quantities, are assumed to be functions of the latter.

If necessary, to model highly accurate measurements or for a specific scope of investigation, themathematical model stated in expression (13) can be refined in a manner suggested in footnotes 2and 3. In contrast to these refinements, a progressive leaning and straightening — neglected to for-mulate the two distances, in expression (4), and resulting in two constant centre-of-mass speeds, V→and Vx — cannot be achieved within this model, as discussed in Appendix C; it is a subject of thesubsequent article (Bos et al, 2020).

The results presented in this article might be an a posteriori reassurance and comfort for Michael Hutchin-son (2006, p. 251), in his attempt to achieve immortality the hard way,

Ride fast—in the end that’s all it ever comes down to. The pressure of another curve,the relief of the simple straight. But the straight’s short respite is never enough. Myshoulders are awful. My arms hurt. And every few seconds I have to manage anotherbanked curve. Each one demands a little more effort, a little more concentration. As thephysical toll mounts, the balance and rhythm aren’t offering the protection from realitythat they did. I need some sort of reassurance. Some comfort.

The power required to maintain the same speed — with respect to the black line — is less on a curvethan on a straight. In a certain manner, the curves provide a short respite, since — thanks to thecurves — the distance travelled by a centre of mass, within a given time, is shorter than the distancewith respect to the black line.

A Lean angle

Expressions (7) and (20) imply that the lean angle, ϑ , depends only on the centripetal accelerationof the cyclist’s centre of mass and the acceleration of gravity. In other words, the lean angle dependsonly on the centre-of-mass speed and the radius of curvature of the centre-of-mass trajectory, noton the track inclination, θ , even though both the normal force, stated in expression (18), andthe frictional force, in expression (19), do depend on track inclination. However, the θ-dependencecancels out of the centripetal force, stated in expression (20), by a seldom used trigonometric identity.

Given the generality of this result, it would be satisfying to obtain it in a manner that explains it inthe context of physics. To this end, let us analyze the situation from inside the noninertial frame ofthe cyclist. Specifically, we consider the frame comoving with the cyclist around a curve, with thecyclist considered as a point mass. We neglect the additional accelerated motion resulting from therotation of the cyclist about an axis through the centre of mass.

As illustrated in Figure A1 and in contrast to Figure 2, in this noninertial frame, instead of theforces in the horizontal direction summing to the centripetal force, Fcp , the forces in this direction,including the fictitious centrifugal force, Fcf = −Fcp , sum to zero. The centrifugal force must betaken to act at the centre of mass, since otherwise the torque about the centre of mass, stated inexpression (16), would be affected.

9

Figure A1: Force diagram: Noninertial frame

To proceed, we invoke the vector identity,∑τ = RCoM ×

∑F +

∑τCoM ,

where∑

τ is the net torque about an arbitrary point,∑

τCoM is the net torque about the centre ofmass, RCoM is the position vector of the centre of mass, and

∑F is the net force. From this identity

there follows the well-known result that if the net force is zero and the net torque about the centreof mass is zero, the net torque about any other point is also zero. In particular, let us consider thetorque about the point of contact of the tires with the surface,∑

τz = hFg sinϑ− hFcf sin(π

2− ϑ

)= 0 ,

from which — considering the magnitudes — it follows that

tanϑ =FcfFg

=FcpFg

, (A.1)

where expression (A.1) is equivalent to expressions (7) and (20). Expression (A.1) manifestly holdswhether the curve is banked or unbanked.

Physically, from inside the noninertial frame of the cyclist, the value of the lean angle that obtainsfor specific values of speed and radius is the one that makes the gravitational torque balance thecentrifugal torque. In particular, this condition makes no reference to track inclination.

B Harmonic mean

To calculate the average power over a lap, we require total work and time,

W =

∮F (s) ds and T =

∮dt =

∮ds

V (s),

to write

P =W

T=

∮F (s) ds∮

ds

V (s)

.

10

If we consider n segments along which F and V are constant, we write

P =

n∑i=1

Fi ∆si

n∑i=1

∆siVi

1

∆s1

∆s

=

n∑i=1

Fi∆si∆s

n∑i=1

1

Vi

∆si∆s

.

Since in our case, there are only two segments, we write

P =F→

∆s→∆s

+ Fx∆sx∆s

1

V→

∆s→∆s

+1

Vx

∆sx∆s

, (B.1)

where → denotes straights and x denotes banks. The numerator is the distance-weighted arithmeticmean, F . The denominator is reciprocal of the distance-weighted harmonic mean,

〈V 〉 =

2∑i=1

∆si∆s

2∑i=1

1

Vi

∆si∆s

=1

1

V→

∆s→∆s

+1

Vx

∆sx∆s

,

which — for the average centre-of-mass speed — results in expression (12a).

In contrast to 〈V 〉 , the distance-weighted arithmetic mean is

V =

2∑i=1

Vi ∆si

∆s= V→

∆s→∆s

+ Vx∆sx∆s

, (B.2)

which, in view of expression (B.1), confirms that P 6= F V . The harmonic mean is less than thearithmetic mean; it is skewed toward slower speeds. Thus, F V would overestimate the averagepower.

For our numerical example, following expression (12), 〈V 〉 = 11.8091 m/s ; hence, in accordancewith expression (13), F 〈V 〉 = 229.6723 W . On the other hand, following expression (B.2), V =11.8114 m/s , and F V = 229.7161 W . The difference is small due to the similarity of values of V→and Vx . It would not be so, for an average of an upwind and downwind segments, discussed byDanek et al (2020b, Appendix A), where also the harmonic mean is used.

C Transition between curves and straights

In this article, to consider the velodrome in question, we assume that, along the banks, the radiusof curvature is constant. Hence, the track is composed of two semicircles and two straights; this isthe case of the light grey oval in Figure C1. The dark grey and black ovals also represent a trackwhose S = 250 m , but their radii of curvature are not constant; they are r = 23 m , at the interceptswith the horizontal axis, and r →∞ , at the intercepts with the vertical axis. The dark grey oval iscomposed of an ellipse (Benham et al, 2020), whose semiaxes are 23 and 30 , and of two straights.The black oval, in polar coordinates, is

r(φ) = 28.15(1 + 0.5 sin2φ cos2φ+ 0.7 cos4φ

), φ = [ 0, 2π) , (C.1)

11

Figure C1: Velodrome tracks

where the coefficients are found numerically by invoking the concept of the arclength and curvatureto ensure that S = 250 m , rmin = 23 m , rmax →∞ .

These ovals share important geometrical properties, namely, their circumference and their radii ofcurvature, at the horizontal-axis and vertical-axis intercepts. However, the model presented in thisarticle applies explicitly to the light grey oval. Its application to other similar ovals entails a decreasein accuracy.

For the light-grey oval, a model requires two constant centre-of-mass speeds, for the straights andfor the circular banks. For the dark-grey oval, an explicit model requires the centre-of-mass speedto be represented by two functions, where the speed along the straight is constant, but along thebanks is not, due to the changing radius of curvature along the elliptical bank. For the black oval,an explicit model requires the centre-of-mass speed to be represented by a single function, whichdepends on the continuously changing radius of curvature.

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Let us examine the black-oval model. Its average curvature, for the length of a lap, is

κ =

S∫0

κ(s) ds

S∫0

ds

=

2π∫0

∣∣∣∣∣r2(φ) + 2

(∂r(φ)

∂φ

)2

− r(φ)∂2r(φ)

∂φ2

∣∣∣∣∣√r2(φ) +

(∂r(φ)

∂φ

)23

√r2(φ) +

(∂r(φ)

∂φ

)2

dφ

2π∫0

√r2(φ) +

(∂r(φ)

∂φ

)2

dφ

=

2π∫0

∣∣∣∣∣r2(φ) + 2

(∂r(φ)

∂φ

)2

− r(φ)∂2r(φ)

∂φ2

∣∣∣∣∣r2(φ) +

(∂r(φ)

∂φ

)2 dφ

2π∫0

√r2(φ) +

(∂r(φ)

∂φ

)2

dφ

;

a numerical integration, with r given in expression (C.1), results in κ = 0.0251257 m−1 . Hence, theaverage radius of curvature is r := 1/ κ = 39.7998 m . This is consistent with an expectation in viewof the light-grey oval, whose radius of curvature is r = 23 m , along the banks, and infinity along thestraights. For the black-oval model, in a manner analogous to expressions (3) and (4), the distancetraveled — in one lap — by the centre of mass is

2π∫0

√√√√√(r(φ)− h sinϑ)2

+

∂(r(φ)− h sinϑ

)∂φ

2

dφ

=

2π∫0

√(r(φ)− h sinϑ

)2

+

(∂r(φ)

∂φ

)2

dφ ,

where ϑ is the average lean angle. Thus, given a laptime, in a manner analogous to expression (6),we write

t =

2π∫0

√(r(φ)− h sinϑ

)2

+

(∂r(φ)

∂φ

)2

dφ

V, (C.2)

where V is the average speed, which — following expression (9) — we write as

V =√g r tanϑ . (C.3)

To compare the resulting power with the numerical example in Section 3, we let the laptime be suchthat, according to expression (5), the corresponding black-line speed is V→ = 12 m/s . In accordancewith expressions (C.2) and (C.3), and as shown in Figure C2, we obtain ϑ = 0.347284 rad ≡ 19.8979◦ ,numerically, which results in V = 11.8878 m/s . Hence, in accordance with expression (2), P =

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Figure C2: Left-hand and right-hand sides of equation (C.2)

236.415 W .5 If we consider the average centre-of-mass speed, per lap, for the light-grey oval, which —in accordance with expression (12) — is V = 11.8114 m/s , we obtain P = 232.223 W . Anothercomparison is the distance travelled by the centre of mass. For the black oval, it is 247.662 m ; forthe light-grey oval, it is 246.024 m .

Figure C3: Track inclination

The model based on the light-gray oval requires fewer approximations, within the realm of math-ematics. This is a consequence of the idealization of a velodrome track, which is greater for thelight-gray oval than for the black oval, due to the assumption of a constant curvature and no transi-tion between the curves and the straights. Thus, in spite of more mathematical approximations, thelatter might exhibit a superior empirical adequacy. A further examination of this question, whichis essential to the concept of modelling, requires experimental results. The conclusiveness of suchresults, however, might also be questionable, in view of the similarity of P = 236.35 W , for the blackoval, and P = 229.7161 W , for the light-grey oval.

Acknowledgements

We wish to acknowledge Len Bos for his mathematical insights, Roger Mason for his perceptivecomments, David Dalton, for his scientific editing and proofreading, Elena Patarini, for her graphic

5We could refine the black-oval model by including the effect of the track inclination, which — given a minimumand maximum values of inclination along the oval, stated in expression (C.1), as well as an interpolation formulabetween them — is illustrated in Figure C3. Hence, this effect could be expressed as a continuous function of distance.

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support, and Roberto Lauciello, for his artistic contribution. Furthermore, we wish to acknowledgeFavero Electronics for inspiring this study by their technological advances and for supporting thiswork by providing us with their latest model of Assioma Duo power meters.

Conflict of Interest

The authors declare that they have no conflict of interest.

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