On modelling bicycle power for velodromes: Part II Formulation for individual pursuits
Len Bos, Michael A. Slawinski, Raphaël A. Slawinski, Theodore Stanoev
OOn modelling bicycle power for velodromes: Part II
Formulation for individual pursuits
Len Bos ∗ , Michael A. Slawinski † , Rapha¨el A. Slawinski ‡ , Theodore Stanoev § September 21, 2020
Abstract
We model the instantaneous power on a velodrome for individual pursuits, taking into ac-count its straights, circular arcs, and connecting transition curves. The forces opposing themotion are air resistance, rolling resistance, lateral friction and drivetrain resistance. We exam-ine the constant-cadence and constant-power cases, and discuss their results, including changesin the kinetic and potential energy.
This article is a continuation of research presented by Danek et al. (2020a), Danek et al. (2020b), Boset al. (2020) and, in particular, by Slawinski et al. (2020). Herein, using a mathematical model, weexamine the power required on a velodrome, for an individual pursuit. The opposing forces consist ofair resistance, rolling resistance, lateral friction and drivetrain resistance. We consider a velodromewith its straights, circular arcs, and connecting transition curves, whose inclusion — while presentinga certain challenge, and neglected in previous studies (e.g., Slawinski et al., 2020) — increases theempirical adequacy of the model.We begin this article by expressing mathematically the geometry of both the black line and theinclination of the track. Our expressions are accurate analogies for the common geometry of modern250-metre velodromes (Mehdi Kordi, pers. comm., 2020 ). We proceed to formulate an expressionfor power expended against dissipative forces, which we examine for both the constant-cadence andconstant-power cases. For either case, we consider, a posteriori , changes in the kinetic and potentialenergy. We conclude by discussing the results. To model the required power for an individual pursuit of a cyclist who follows the black line, in aconstant aerodynamic position, we define this line by three parameters. ∗ Universit`a di Verona, Italy, [email protected] † Memorial University of Newfoundland, Canada, [email protected] ‡ Mount Royal University, Canada, [email protected] § Memorial University of Newfoundland, Canada, [email protected] The circumference along the inner edge of this five-centimetre-wide line — also known as the measurement lineand the datum line — corresponds to the official length of the track. a r X i v : . [ phy s i c s . a pp - ph ] S e p L s : the half-length of the straight– L t : the length of the transition curve between the straight and the circular arc– L a : the half-length of the circular arcThe length of the track is S = 4( L s + L t + L a ) . In Figure 1, we show a quarter of a black line for L s = 19 m , L t = 13 . L a = 30 m , which results in S = 250 m . This curve has continuous Figure 1:
A quarter of the black line for a 250-metre track derivative up to order two; it is a C curve, whose curvature is continuous.To formulate, in Cartesian coordinates, the curve shown in Figure 1, we consider the following.– The straight, y = 0 , (cid:54) x (cid:54) a , shown in gray, where a := L s .– The transition, shown in black — following a standard design practice — we take to be an Eulerspiral, which can be parameterized by Fresnel integrals, x ( ς ) = a + (cid:114) A ς √ A (cid:90) cos (cid:0) x (cid:1) d x and y ( ς ) = (cid:114) A ς √ A (cid:90) sin (cid:0) x (cid:1) d x , with A > ς is a curve parameter. Since the arclength differen-tial, d s , is such that d s = (cid:113) x (cid:48) ( ς ) + y (cid:48) ( ς ) d ς = (cid:115) cos (cid:18) Aς (cid:19) + sin (cid:18) Aς (cid:19) d ς = d ς , we write the transition curve as( x ( s ) , y ( s )) , (cid:54) s (cid:54) b := L t . c , c ) and whose radius is R , with c , c and R to be determined. Since its arclength is specified to be c := L a , we may parameterizethe quarter circle by x ( θ ) = c + R cos( θ ) (1)and y ( θ ) = c + R sin( θ ) , (2)where − θ (cid:54) θ (cid:54) θ := c/R . The centre of the circle is shown as a black dot in Figure 1.We wish to connect these three curve segments so that the resulting global curve is continuous alongwith its first and second derivatives. This ensures that the curvature of the track is also continuous.To do so, let us consider the connection between the straight and the Euler spiral. Herein, x (0) = a and y (0) = 0 , so the spiral connects continuously to the end of the straight at ( a,
0) . Also, at( a,
0) , d y d x = y (cid:48) (0) x (cid:48) (0) = 01 = 0 , which matches the derivative of the straight line. Furthermore, the second derivatives match, sinced y d x = y (cid:48)(cid:48) (0) x (cid:48) (0) − y (cid:48) (0) x (cid:48)(cid:48) (0)( x (cid:48) (0)) = 0 , which follows, for any A > x (cid:48) ( ς ) = cos (cid:18) A ς (cid:19) , y (cid:48) ( ς ) = sin (cid:18) A ς (cid:19) (3)and x (cid:48)(cid:48) ( ς ) = − A ς sin (cid:18)
A ς (cid:19) , y (cid:48)(cid:48) ( ς ) = A ς cos (cid:18)
A ς (cid:19) . Let us consider the connection between the Euler spiral and the arc of the circle. In order that theseconnect continuously, (cid:0) x ( b ) , y ( b ) (cid:1) = (cid:0) x ( − θ ) , y ( − θ ) (cid:1) , we require x ( b ) = c + R cos( θ ) ⇐⇒ c = x ( b ) − R cos (cid:16) cR (cid:17) (4)and y ( b ) = c − R sin( θ ) ⇐⇒ c = y ( b ) + R sin (cid:16) cR (cid:17) . (5)For the tangents to connect continuously, we invoke expression (3) to write( x (cid:48) ( b ) , y (cid:48) ( b )) = (cid:18) cos (cid:18) A b (cid:19) , sin (cid:18) A b (cid:19)(cid:19) . Following expressions (1) and (2), we obtain (cid:0) x (cid:48) ( − θ ) , y (cid:48) ( − θ ) (cid:1) = (cid:0) R sin( θ ) , R cos( θ ) (cid:1) , respectively. Matching the unit tangent vectors results incos (cid:18) A b (cid:19) = sin (cid:16) cR (cid:17) , sin (cid:18) A b (cid:19) = cos (cid:16) cR (cid:17) . (6)3or the second derivative, it is equivalent — and easier — to match the curvature. For the Eulerspiral, κ ( s ) = x (cid:48) ( s ) y (cid:48)(cid:48) ( s ) − y (cid:48) ( s ) x (cid:48)(cid:48) ( s ) (cid:16)(cid:0) x (cid:48) ( s ) (cid:1) + (cid:0) y (cid:48) ( s ) (cid:1) (cid:17) = A s cos (cid:18) A s (cid:19) + A s sin (cid:18) A s (cid:19) = A s , which is indeed the defining characteristic of an Euler spiral: the curvature grows linearly in thearclength. Hence, to match the curvature of the circle at the connection, we require
A b = 1 R ⇐⇒ A = 1 b R . Substituting this value of A in equations (6), we obtaincos (cid:18) b R (cid:19) = sin (cid:16) cR (cid:17) , sin (cid:18) b R (cid:19) = cos (cid:16) cR (cid:17) ⇐⇒ b R = π − cR ⇐⇒ R = b + 2 cπ . It follows that A = 1 b R = πb ( b + 2 c ) ;hence, the continuity condition stated in expressions (4) and (5) determines the centre of the cir-cle, ( c , c ) .For the case shown in Figure 1, the numerical values are A = 3 . × − m − , R = 23 . c = 25 . c = 23 . ,
0) — isshown in Figure 2. The corresponding curvature is shown in Figure 3. Note that the curvature
Figure 2:
Black line of 250-metre track transitions linearly from the constant value of straight, κ = 0 , to the constant value of the circulararc, κ = 1 /R . 4 igure 3: Curvature of the black line, κ , as a function of distance, s , with a linear transition between thezero curvature of the straight and the 1 /R curvature of the circular arc Figure 4:
Track inclination, θ , as a function of the black-line distance, s There are many possibilities to model the track inclination angle. We choose a trigonometric formulain terms of arclength, which is a good analogy of an actual 250-metre velodrome. The minimuminclination of 13 ◦ corresponds to the midpoint of the straight, and the maximum of 44 ◦ to the apexof the circular arc. For a track of length S , θ ( s ) = 28 . − . (cid:18) πS s (cid:19) ; (7) s = 0 refers to the midpoint of the lower straight, in Figure 2, and the track is oriented in thecounterclockwise direction. Figure 4 shows this inclination for S = 250 m . A mathematical model to account for the power required to propel a bicycle is based on (e.g., Daneket al., 2020a) P = F V , where F stands for the magnitude of forces opposing the motion and V for speed. Herein, we modelthe rider as undergoing instantaneous circular motion, in rotational equilibrium about the line ofcontact of the tires with the ground. Following Slawinski et al. (2020, Section 2), in accordance with5igure 5, along the black line of a velodrome, in windless conditions, P =11 − λ (cid:40) (8a) (cid:32) C rr F g (cid:122)(cid:125)(cid:124)(cid:123) m g (sin θ tan ϑ + cos θ ) (cid:124) (cid:123)(cid:122) (cid:125) N cos θ + C sr (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F g (cid:122)(cid:125)(cid:124)(cid:123) m g sin( θ − ϑ )cos ϑ (cid:124) (cid:123)(cid:122) (cid:125) F f (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sin θ (cid:33) v (8b)+ C d A ρ V (cid:41) , (8c)where m is the mass of the cyclist and the bicycle, g is the acceleration due to gravity, θ is thetrack-inclination angle, ϑ is the bicycle-cyclist lean angle, C rr is the rolling-resistance coefficient,C sr is the coefficient of the lateral friction, C d A is the air-resistance coefficient, ρ is the air density, λ is the drivetrain-resistance coefficient. Herein, v is the speed at which the contact point of therotating wheels moves along the track, which is equivalent to the black-line speed (Danek et al.,2020a, Appendix B), and V is the centre-of-mass speed. Since the lateral friction is a dissipativeforce, it does negative work, and the work done against it — as well as the power — are positive. Forthis reason, in expression (8b), we consider the magnitude, (cid:12)(cid:12) (cid:12)(cid:12) . Figure 5:
Force diagram
In expression (8), we assume the steadiness of effort, which — following an initial acceleration — isconsistent with a steady pace of an individual pursuit. Formally, this assumption corresponds tosetting the acceleration, a , to zero in Slawinski et al. (2020, expression (1)). Herein, the accelerationrefers to the change of the centre-of-mass speed. This speed is nearly constant if the power isconstant, which can be viewed as a quantification of the cyclist’s effort. In other words, the force —and hence the power required to accelerate the bicycle-cyclist system — is associated mainly withthe change of the centre-of-mass speed, not with the change of the black-line speed.To gain an insight into expression (8), let us consider a few special cases. If θ = ϑ = 0 , P = C rr m g + C d A ρ V − λ (cid:124) (cid:123)(cid:122) (cid:125) F V , (9)where — as expected for a flat, straight road — v ≡ V . Also, on a velodrome, along the straights, ϑ = 0 and expression (8b) becomes (cid:0) C rr m g cos θ + C sr m g sin θ (cid:1) V . ϑ = θ , the second summand of expression (8b) is zero, as expected.Let us return to expression (8). Therein, θ is given by expression (7). The lean angle is (Slawinskiet al., 2020, Appendix A) ϑ = arctan V g r CoM , (10)where r CoM is the centre-of-mass radius, and — along the curves, at any instant — the centre-of-massspeed is V = v r CoM (cid:122) (cid:125)(cid:124) (cid:123) ( R − h sin ϑ ) R = v (cid:18) − h sin ϑR (cid:19) , (11)where R is the radius discussed in Section 2.1 and h is the centre-of-mass height. Along the straights,the black-line speed is equivalent to the centre-of-mass speed, v = V . As expected, V = v if h = 0 , ϑ = 0 or R = ∞ .Invoking expressions (10) and (11), we neglect the vertical variation of the centre of mass and, hence,assume that the centre-of-mass trajectory is contained in a horizontal plane, where — in accordancewith the track geometry — this plane is parallel to the plane that contains the black line. Accountingfor the vertical motion of the centre of mass would mean allowing for a nonhorizontal centripetalforce and including the work done in raising the centre of mass. For expressions (8), (10) and (11), we consider a velodrome discussed in Section 2, to let R =23 . h = 1 . m = 84 kg , C d A = 0 . , C rr =0 .
002 , C sr = 0 .
003 and λ = 0 .
02 . For the external conditions, g = 9 .
81 m/s and ρ = 1 .
225 kg/m . Let the black-line speed be constant, v = 16 . V = 16 . V , shown in Figure 7, result from the lean angle. Along thestraights, ϑ = 0 = ⇒ V = v . Along the curves, since ϑ (cid:54) = 0 , the centre-of-mass travels along ashorter path; hence, V < v . Thus, assuming a constant black-line speed implies a variable centre-of-mass speed and, hence, an acceleration and deceleration, even though d
V / d t , where t stands fortime, is not included explicitly in expression (8). Examining Figure 7, we conclude that d V / d t (cid:54) = 0along the transition curves only.The power — obtained by evaluating expression (8), at each point along the track — is shown inFigure 8. The average power, per lap, is P = 580 . igure 6: Lean angle, ϑ , as a function of the black-line distance, s , for constant cadence Figure 7:
Centre-of-mass speed, V , as a function of the black-line distance, s , for constant cadence Examining Figure 8, we see the decrease of power required to maintain the same black-line speedalong the curve. This is due to both the decrease of the centre-of-mass speed, which results ina smaller value of term (8c), and the decrease of a difference between the track-inclination angleand the lean angle, shown in Figure 9, which results in a smaller value of the second summand ofterm (8b).The argument presented in the previous paragraph leads to the following conjecture. The mostefficient track is circular with θ = ϑ , which would correspond to the dashed line in Figure 9. However,this is not possible, since — according to the regulations of the Union Cycliste Internationale — theinner edge of the track shall consist of two curves connected by two parallel straight lines. Hence,the optimization is constrained by the length of the straights.Examining Figure 10, where — in accordance with expression (8) — we distinguish among the powerused to overcome the air resistance, the rolling resistance and the lateral friction, we can quantifytheir effects. The first has the most effect; the last has the least effect, and is zero at points forwhich θ = ϑ , which corresponds to the zero crossings in Figure 9.Let us comment on potential simplifications of a model. If we assume a straight flat course — whichis tantamount to neglecting the lean and inclination angles — we obtain, following expression (9), P ≈
610 W . If we consider an oval track but ignore the transitions and assume that the straightsare flat and the semicircular segments, whose radius is 23 m , have a constant inclination of 43 ◦ ,we obtain (Slawinski et al., 2020, expression (13)) P ≈
563 W . In both cases, there is a significantdiscrepancy with the power obtained from the model discussed herein, P = 573 . t − t , is W = t (cid:90) t P d t = 1 v s (cid:90) s P v d t (cid:124)(cid:123)(cid:122)(cid:125) d s , where the black-line speed, v , is constant and, hence, d s is an arclength distance along the black8 igure 8: Power, P , as a function of the black-line distance, s , for constant cadence Figure 9: θ − ϑ , as a function of the black-line distance, s , for constant cadence line. Considering the average power per lap, we write W = Sv (cid:124)(cid:123)(cid:122)(cid:125) t (cid:9) S (cid:82) P d sS (cid:124) (cid:123)(cid:122) (cid:125) P = P t (cid:9) . Given P = 580 . t (cid:9) = 14 . W = 8691 . Let us solve numerically the system of nonlinear equations given by expressions (8), (10) and (11),to find the lean angle as well as both speeds, v and V , at each point of a discretized model of thetrack , under the assumption of constant power. In accordance with a discussion in Section 3, suchan assumption is more consistent with the steadiness of effort than the assumption of a constantcadence examined in Section 4.2.As in Section 4.2, we let R = 23 . h = 1 . m = 84 kg , C d A = 0 . , C rr = 0 .
002 ,C sr = 0 .
003 , λ = 0 .
02 , g = 9 .
81 m/s and ρ = 1 .
225 kg/m . However, in contrast to Section 4.2,we allow the black-line speed to vary, and set the power to be the average obtained in that section, P = 580 . v = V RR − h sin ϑ , igure 10: Power to overcome air resistance, rolling resistance and lateral friction we write expression (8) as P = (12) V − λ (cid:40)(cid:32) C rr m g (sin θ tan ϑ + cos θ ) cos θ + C sr (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m g sin( θ − ϑ )cos ϑ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sin θ (cid:33) RR − h sin ϑ + C d A ρ V (cid:41) , where, in agreement with expression (10), ϑ = arctan V g ( R − h sin ϑ ) , which — given g , R and h — can be solved for V as a function of ϑ . Inserting that solution inexpression (12), we obtain an equation whose only unknown is ϑ .The difference of the lean angle — between the case of a constant cadence and a constant power isso small that there is no need to plot it; Figure 6 illustrates it accurately. The same is true for thedifference between the track-inclination angle and the lean angle, illustrated in Figure 9, as well asfor the dominant effect of the air resistance, illustrated in Figure 10.The resulting values of V are shown in Figure 11. As expected, in view of the dominant effect of theair resistance, a constancy of P entails small variations in V . In comparison to the case discussedin Section 4.2, the case in question entails lesser accelerations and decelerations of the centre ofmass — note the difference of vertical scale between Figures 7 and 11 — but the changes of speedare not limited to the transition curves. Even though such changes are not included explicitly inexpression (8), a portion of the given power may be accounted for by m V d V / d t , which is associatedwith accelerations and decelerations. The amount of this portion can be estimated a posteriori .10ince m V d V d t = dd t (cid:18) m V (cid:19) , the time integral of the power used for acceleration of the centre of mass is the change of its kineticenergy. Therefore, to include the effect of accelerations, per lap, we need to add the increases in ki-netic energy. This is an estimate of the error committed by neglecting accelerations in expression (8)to be quantified, for the constant-cadence and constant-power cases, in Section 5.The values of v , in accordance with expression (11), are shown in Figure 12. The averages are V = 16 . v = 16 . . . t − t , is W = t (cid:90) t P d t = P t (cid:90) t d t = P ( t − t ) (cid:124) (cid:123)(cid:122) (cid:125) t (cid:9) = P t (cid:9) , where, for the second equality sign, we use the constancy of P ; also, we let the time interval to bea laptime. Thus, given P = 580 . t (cid:9) = 14 . W = 8689 . Figure 11:
Centre-of-mass speed, V , as a function of the black-line distance, s , for constant power Figure 12:
Black-line speed, v , as a function of the black-line distance, s , for constant power In Section 4, we calculate the power expended and work done by the cyclist against dissipativeforces, namely, rolling friction, lateral friction and air resistance, while neglecting changes in the11yclist’s kinetic and potential energy. However, even assuming a constant black-line speed togetherwith a black line contained within a horizontal plane, the speed and height of the centre of masschange along the curved segments of the track, resulting in changes in kinetic and potential energy.In this section, we calculate the work the cyclist performs to effect these changes. In the caseof a constant black-line speed, discussed in Section 4.2, the work done to increase the system’smechanical energy can be simply added to the work done against the dissipative forces to find thetotal work done over a single lap. In the case of constant power, discussed in Section 4.3, however,the work done to increase the system’s mechanical energy cannot be so added, but only estimatedapproximately. This is because, strictly speaking, including that work contradicts the assumption ofconstant power. To include — in a rigorous manner — the work done to increase mechanical energyin the constant-power case would require modifying the model for instantaneous power, stated inexpression (8). Still, we can use an a posteriori calculation to estimate the relative importance ofincreasing mechanical energy in that case.Consider a cyclist following a track whose curvature is R , with a centre-of-mass speed, V . Wewish to determine the work the cyclist must perform to change the kinetic and potential energywhen the radius of curvature and speed change to new values, R and V , respectively. Let us firstdetermine the work required to change the kinetic energy.Neglecting the kinetic energy of the rotating wheels, the kinetic energy of the system is due to thetranslational motion of the centre of mass, K = m V . To find the work done to change kinetic energy, it is important to note that the cyclist-bicycle systemis not purely mechanical, in the sense that in addition to kinetic and potential energy the systemalso possesses internal energy in the form of the cyclist’s chemical energy stores. When the cyclistspeeds up, internal energy is converted into kinetic. However, the converse is not true: when thecyclist slows down, kinetic energy is not converted into internal. It follows that, to determine thework the cyclist does to change kinetic energy, we should consider only the increases. Then, if thecentre-of-mass speed increases monotonically from V to V , the work done is simply the increasein kinetic energy, ∆ K = m (cid:0) V − V (cid:1) . The stipulation of a monotonic increase is needed due to the non-mechanical nature of the system,with the cyclist doing positive work — with an associated decrease in internal energy — when speed-ing up, but not doing negative work — with an associated increase in internal energy — when slowingdown. Put another way, the cyclist’s work to increase speed from V to V depends not only on theinitial and final speeds but also on the intermediate speeds.Now, let us determine the work required to change the potential energy, U = m g h cos ϑ . The lean angle , ϑ , is determined by assuming that, at all times, the system is in rotational equi-librium about the line of contact of the tires with the ground; this assumption yields the implicitcondition on ϑ , stated in expression (10). Since the lean angle depends on the centre-of-massspeed, V , as well as the radius of curvature of the track, R , it, and therefore the height of the centreof mass, change if either V or R changes. The work done is simply the increase in potential energy U resulting from a monotonic decrease in the lean angle from ϑ to ϑ ,∆ U = m g h (cos ϑ − cos ϑ ) , as can be seen in Figure 6. The same considerations due to the nonmechanical nature of the systemapply to the work done to change potential energy as to that done to change kinetic energy. The12yclist does positive work — with an associated decrease in internal energy — when straightening up,but not negative work — with an associated increase in internal energy — when leaning into the turn.And as before, the cyclist’s work to decrease in the lean angle from ϑ to ϑ depends not only onthe initial and final lean angles but also on the intermediate angles.Note that the changes in potential energy due to a changing lean angle are different in character fromthe changes associated with hill climbs, stated as the first term in the numerator of expression (1)of Slawinski et al. (2020). When climbing a hill, a cyclist does work to increase potential energy.However, when descending the hill, at least some of that potential energy is converted into kineticenergy of forward motion. This is not the case with the work the cyclist does to straighten up. Whenthe cyclist leans again, potential energy is not converted into kinetic energy of forward motion. In the example considered in Section 4.2, assuming a constant cadence, which is equivalent to aconstant black-line speed, the sum of increases of the centre-of-mass speed squared over one lap,shown by thick black lines in Figure 13, is (cid:80) ∆ V = 42 . / s . This results an increase inkinetic energy of ∆ K = m (cid:88) ∆ V = 1789 . . In the same example, the sum of increases of the centre-of-mass height over one lap is (cid:80) h ∆ cos ϑ =0 . U = m g h (cid:88) ∆ cos ϑ = 691 . . Adding these to the work done against dissipative forces calculated in Section 4.2, we find that thetotal work done by the cyclist over one lap is W = 8691 . . . . . Figure 13:
Increases of V , as a function of the black-line distance, s , for constant cadence In the example considered in Section 4.3, we assume a constant instantaneous power needed toovercome dissipative forces, without including changes in mechanical energy. The sum of increasesof the centre-of-mass speed squared over one lap, shown by thick black lines in Figure 14, is (cid:80) ∆ V =20 . / s , which results in an increase in kinetic energy of∆ K = m (cid:88) ∆ V = 880 . . igure 14: Increases of V , as a function of the black-line distance, s , for constant power In the same example, the sum of increases of the centre-of-mass height over one lap is (cid:80) h ∆ cos ϑ =0 . U = m g h (cid:88) ∆ cos ϑ = 716 . . Adding these to the work done against dissipative forces calculated in Section 4.3, we find that thetotal work done by the cyclist over one lap is W = 8689 . . . . . The mathematical model presented in this article offers the basis for a quantitative study of individ-ual pursuits. The model can be used to predict or retrodict the laptimes, from the measurementsof power, or to estimate the power from the recorded times. Comparisons of such predictions orretrodictions with the measurements of time, speed, cadence and power along the track offer aninsight into the empirical adequacy of a model. Given a satisfactory adequacy and appropriate mea-surements, the model lends itself to estimating the rolling-resistance, lateral-friction, air-resistanceand drivetrain-resistance coefficients.Presented results allow us to comment on aspects of the velodrome design. As illustrated in Fig-ures 6–8, 11, 12, the transitions — between the straights and the circular arcs — are not smooth forthe lean angles, speeds and powers. It might suggest that a commonly used Euler spiral, illustratedin Figure 3, is not the optimal transition curve. Perhaps, the choice of a transition curve shouldconsider such phenomena as the jolt, which is the temporal rate of change of acceleration. It mightalso suggest the necessity for the lengthening of the transition curve.Furthermore, an optimal velodrome design would strive to minimize the distance between the zeroline and the curve in Figure 9, which is tantamount to optimizing the track inclination to accom-modate the lean angle of a rider. The smaller the distance, the smaller the second summand interm (8b). As the distance tends to zero, so does the summand.These considerations are to be explored more thoroughly in a future work. Also, the inclusion ofchange of kinetic and potential energy within the model for instantaneous power is a refinement tobe considered. 14 cknowledgements
We wish to acknowledge Mehdi Kordi, for insights into the track geometry, Elena Patarini, for hergraphic support, and Favero Electronics for inspiring this study by their technological advances withtheir latest model of Assioma Duo power meters.
Conflict of Interest
The authors declare that they have no conflict of interest.
References
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