On minimizing cyclists’ ascent times: Part II

Abstract

We formulate an optimization of a bicycle ascent time under the constraints of the average, maximum, and minimum powers. In contrast to the first part of this study, we do not restrict the departure to flying starts with an initial speed determined by the model and its optimization. We allow for various initial speeds, from a standstill to a launched start. We accomplish this by generalizing the discontinuous piecewise constant speed model to a continuous piecewise linear speed model. Regardless of the initial speed, steepness, or profile of the ascent, the optimal strategy tends to a constant ground speed, in agreement with the conclusion of the previous, more restricted, formulation. This new formulation allows us to compare various initial-speed strategies and, hence, has a direct application to competitive cycling. Notably, in timetrials composed of flat and steep sections, it helps one decide whether or not to change the bicycle, which requires stopping and restarting, from one that is more appropriate for flats to one that is more appropriate for uphills.

1. Introduction

In this paper, we elaborate on a strategy proposed by Bos et al. [1–3], which could be viewed as Part I of the present work, to minimize the ascent time on a bicycle, which is the goal of timetrials that include hill climbs. In a manner similar to Bos et al. [3], we assume that the ascent, including distance and steepness, is known, and so is the weight of the bicycle-cyclist system, as well as fitness level, quantified by the power output, which—for the last quarter of the century—has been the key metric in training and race preparation. Other known information consists of the air, rolling, and drivetrain resistances.

In contrast to Bos et al. [1–3], we do not restrict our study to flying starts with an initial speed determined by the model and its optimization but allow different initial speeds, including a standstill departure and a launched start, which are the two cases exemplified in this paper. We accomplish this by generalizing the discontinuous piecewise constant speed model of the previous paper to a continuous piecewise linear speed model. This modification, which renders the model more flexible and empirically adequate, entails challenges in its formulation and subsequent computations.

Several other authors used similar mathematical models to conduct studies on uphill timetrialing. Atkinson et al. [4] analyze variable versus constant power strategies with consideration of the cyclist power output, gradient of an ascent, and wind velocity for three race courses of varying lengths. Boswell [5] considers a differential equation model under the constraint of a constant-average power and considers various pacing strategies to find that substantial time savings can be realized by cyclists increasing their power on uphill sections and suitably reducing their work rates elsewhere. Also, Boswell [5] includes a derivation of an optimal pacing strategy, subject to a constant average power, to be a constant speed, which is similar to the findings of Bos et al. [1–3] but does not consider a maximum power. Roa and Muñoz [6] develop a methodology to analyze the bicycle change strategy for hilly timetrial races, which involves solving an optimization problem to minimize the race time of a predetermined route as a function of the bicycle-change location along the route. Feng et al. [7] incorporate empirical fatigue power profiles of professional cyclists and, based on the principles of mechanics and biological fatigue, solve optimization problems for the shortest competition finishing time implemented on the men 2021 UCI and women 2021 Olympic individual timetrial courses.

We begin this paper by invoking a phenomenological model to estimate the power required for a given speed. This model and its notation are similar to the ones used and discussed in several previous studies by Bos et al. [1–3,8,9]. Subsequently, we examine a numerical solution of the ascent-time minimization, and exemplify this solution for both standstill departures and launched starts. We conclude the paper by commenting on advancements with respect to the formulation of Bos et al. [1–3] and on the application of this work to competitive cycling.

2. Phenomenological power model

2.1. Physical considerations

We consider a phenomenological model for a bicycle-cyclist system of mass m moving with ground speed V. ¹ The effort expended by the cyclist is quantified by power $P=P_K+P_F$ ; the former is required to change the kinetic energy, $\frac{1}{2}m{V}^2$ , and the latter to overcome opposing forces, with both affected by the drivetrain resistance. Hence,

$$P=\frac{\stackrel{P_K}{\overbrace{m\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}t}V}}+\stackrel{P_F}{\overbrace{FV}}}{1-\lambda },$$

(1)

where λ is the drivetrain-resistance coefficient. $P_F$ is affected by change in elevation, as well as the resistance of rolling and of air. Thus, explicitly,

$$\begin{array}{ccc}\hfill P& =\frac{\stackrel{\mathrm{change}\mathrm{in}\mathrm{speed}}{\overbrace{m\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}t}}}+\stackrel{\mathrm{change}\mathrm{in}\mathrm{elevation}}{\overbrace{mg\sin \theta }}+\stackrel{\mathrm{rolling}\mathrm{resistance}}{\overbrace{{\mathrm{C}}_{\mathrm{rr}}\underset{\mathrm{normal}\mathrm{force}}{\underbrace{mg\cos \theta }}}}+\stackrel{\mathrm{air}\mathrm{resistance}}{\overbrace{\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}{V}^2}}}{\underset{\mathrm{drivetrain}\mathrm{efficiency}}{\underbrace{1-\lambda }}}V,\hfill & \hfill \end{array}$$

(2)

where g is the acceleration due to gravity, θ is the slope angle, and ρ is the air density. Thus, the three model parameters are the aforementioned λ, together with the rolling-resistance coefficient, ${\mathrm{C}}_{\mathrm{rr}}$ , and the air-resistance coefficient, ${\mathrm{C}}_{\mathrm{d}}\mathrm{A}$ .

2.2. Power-model discretization

In previous studies, Bos et al. [1–3] sought to determine the optimal strategy for minimizing the ascent time subject to the constraint of a given average power. The hill was assumed to have no flat or descent sections and was modeled by N straight-line segments. Each segment was traversed at a constant speed with discontinuous speed between segments. Herein, the hill exhibits the same characteristics, but the speed is represented by a piecewise linear interpolation of values at the endpoints of each segment. This assumption corresponds to assuming a constant acceleration on each segment.

Bos et al. [3] neglected the first term in the numerator of expression (2), which is the power, $P_K$ , associated with changes in kinetic energy. This was justified by the fact that, riding uphill, speeds are typically low, and changes in kinetic energy are much smaller than the work done against opposing forces, especially against gravity. Furthermore, Bos et al. [3] restricted their study to flying starts with the speed determined by the minimization of ascent time. Under these conditions, they proved that, given an average power, the ascent time is minimized if a cyclist maintains a constant ground speed regardless of the slope. Herein, we allow for a variable start speed, regardless of the speed that minimizes the subsequent ascent time. Consequently, we include the first term in the numerator of expression (2). As it turns out, following various start speeds, an optimal strategy still tends to a constant speed.

In accordance with expression (2), the power model for the cyclist can be written as

$$P=\frac{1}{1-\lambda }\left(m\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}t}V+\mathrm{FV}\right).$$

(3)

Therefore, the total work done is

$$W=\underset{0}{\overset{T}{\int }}P\mathrm{d}t=\frac{1}{1-\lambda }\left({\frac{1}{2}m{V}^2|}_0^T+\underset{0}{\overset{L}{\int }}F\mathrm{d}s\right),$$

(4)

where the first term in parentheses represents the change in kinetic energy. Note that the remaining integral contains $\mathrm{d}s=V\mathrm{d}t$ .

Assuming the cyclist starts from rest, $V_1=0$ , we obtain

$$W=\frac{1}{1-\lambda }\left(\frac{1}{2}m{V}_T^2+\underset{0}{\overset{L}{\int }}F\mathrm{d}s\right).$$

(5)

Computationally, discretizing the case of $V_1=0$ is the most challenging due to the resulting discontinuity; hence, below we present it explicitly. Computational results for other initial speeds can be obtained in a similar manner without such challenges, as we demonstrate in Section 3.

Within the integrand of expression (5), the increment of the work done against forces opposing the motion is

$$F\mathrm{d}s=\left(mg{\mathrm{C}}_{\mathrm{rr}}\cos \theta +mg\sin \theta +\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}{V}^2\right)\mathrm{d}s.$$

(6)

Since $\cos \theta \mathrm{d}s=\mathrm{d}x$ , the horizontal path increment, and $\sin \theta \mathrm{d}s=\mathrm{d}y$ , the vertical path increment, the contributions to work from opposing gravity and rolling friction are integrable,

$$\begin{array}{cccc}\hfill \underset{0}{\overset{L}{\int }}F\mathrm{d}s& =\underset{0}{\overset{\mathrm{\Delta }X}{\int }}mg{\mathrm{C}}_{\mathrm{rr}}\mathrm{d}x+\underset{0}{\overset{\mathrm{\Delta }Y}{\int }}mg\mathrm{d}y+\underset{0}{\overset{L}{\int }}\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}{V}^2\mathrm{d}s\hfill & \hfill & \hfill \\ \hfill & =mg{\mathrm{C}}_{\mathrm{rr}}\mathrm{\Delta }X+mg\mathrm{\Delta }Y+\underset{0}{\overset{L}{\int }}\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}{V}^2\mathrm{d}s.\hfill & \hfill \end{array}$$

(7)

The only non-integrable contribution to work comes from opposing air resistance.

To calculate this contribution, we discretize the path into N straight-line segments with lengths $\{L_1,\dots ,L_N\}$ and corresponding speeds $\{V_1,\dots ,V_{N+1}\}$ , where $V_j$ and $V_{j+1}$ are at the beginning and the end of the jth segment, respectively. Along the jth segment, we assume that the change in speed is continuous and piecewise linear in t,

$$V(t):=V_j+M_j\left(t-t_j\right),t_j\le t<t_{j+1},\mathrm{for}j=1,\dots ,N,$$

(8)

where $M_j:=(V_{j+1}-V_j)∕(t_{j+1}-t_j)$ is the constant acceleration on the segment, and $t_j$ and $t_{j+1}$ are the ascent times at the endpoints of the jth segment. As a consequence of this assumption, we obtain the distance traveled on the jth segment—whose endpoints are $s_j$ and $s_{j+1}$ and length is $L_j:=s_{j+1}-s_j$ —by integrating

$$\underset{s_j}{\overset{s}{\int }}\mathrm{d}\sigma =\underset{t_j}{\overset{t}{\int }}V(\tau )\mathrm{d}\tau ,$$

(9)

where σ and τ are the integration variables in place of arc length and time, respectively. Substituting speed (8) and evaluating the integrals, we solve for the distance traveled and define

$$s(t):=s_j+\frac{1}{2}\left(t-t_j\right)\left(M_j\left(t-t_j\right)+2V_j\right),t_j\le t<t_{j+1},s_j\le s<s_{j+1},\mathrm{for}j=1,\dots ,N.$$

(10)

Evaluating expression (10) at the end of the jth segment, we obtain $s_{j+1}:=s(t_{j+1})=s_j+\frac{1}{2}\left(V_{j+1}+V_j\right)$ $\left(t_{j+1}-t_j\right)$ , which we rearrange to obtain the change in time over the jth segment as a function of distance

$$t_{j+1}-t_j=\frac{s_{j+1}-s_j}{\frac{1}{2}\left(V_{j+1}+V_j\right)}=\frac{L_j}{V_{(j)}},$$

(11)

where $V_{(j)}$ denotes the discretization of V over the jth segment

$$V_{(j)}:=\frac{V_j+V_{j+1}}{2}.$$

(12)

Also, with expression (11), the acceleration on the jth segment can be expressed with respect to change in t as well as change in s by

$$M_j=\frac{V_{j+1}-V_j}{t_{j+1}-t_j}=\frac{(V_{j+1}-V_j)(V_{j+1}+V_j)}{2L_j}=\frac{V_{j+1}^2-V_j^2}{2L_j}.$$

(13)

Now, returning to the expression (7), we evaluate the remaining integral as a summation along the jth segments of the path using speed (8). Considering only the integrable factors of the term, as $\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}$ is a constant, and reverting the variable of integration to time with $\mathrm{d}s=V\mathrm{d}t$ , we have

$$\underset{0}{\overset{L}{\int }}{V}^2\mathrm{d}s=\underset{0}{\overset{T}{\int }}{V}^3\mathrm{d}t=\sum _{j=1}^N\left(\underset{t_j}{\overset{t_{j+1}}{\int }}V{(t)}^3\mathrm{d}t\right).$$

(14)

Proceeding with u-substitution for the summand, we let $u=V(t)$ and $\mathrm{d}u=V^{\prime }(t)\mathrm{d}t=M_j\mathrm{d}t$ , with $u(t_j)=V(t_j)=V_j$ and $u(t_{j+1})=V(t_{j+1})=V_{j+1}$ . As such

$$\underset{t_j}{\overset{t_{j+1}}{\int }}V{(t)}^3\mathrm{d}t=\underset{V_j}{\overset{V_{j+1}}{\int }}{u}^3\frac{\mathrm{d}u}{M_j}=\frac{1}{M_j}{\left[\frac{u^4}{4}\right]}_{V_j}^{V_{j+1}}=\frac{1}{M_j}\frac{V_{j+1}^4-V_j^4}{4}.$$

(15)

However, in view of slope (13), expressed in terms of the jth segment distance, we simplify the result of integral (15) as

$$\frac{1}{M_j}\frac{V_{j+1}^4-V_j^4}{4}=\frac{2L_j}{V_{j+1}^2-V_j^2}\frac{\left(V_{j+1}^2-V_j^2\right)\left(V_{j+1}^2+V_j^2\right)}{4}=L_j{V}_{(j)}^2,$$

(16)

where $V_{(j)}^2$ denotes the discretization of $V^2$ over the jth segment,

$$V_{(j)}^2:=\frac{V_j^2+V_{j+1}^2}{2}.$$

(17)

Therefore, the work done against air resistance may be written as

$$\underset{0}{\overset{L}{\int }}\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}{V}^2\mathrm{d}s=\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}\sum _{j=1}^N{L}_j{V}_{(j)}^2.$$

(18)

Then, in accordance with expression (4), the total work done—in its discretized form—is

$$W=\frac{\alpha }{2}{V}_{N+1}^2+\mathrm{\alpha g}\left({\mathrm{C}}_{\mathrm{rr}}\mathrm{\Delta }X+\mathrm{\Delta }Y\right)+\beta \sum _{j=1}^N{L}_j{V}_{(j)}^2,$$

(19)

where

$$\alpha :=\frac{m}{1-\lambda }\mathrm{and}\beta :=\frac{\frac{1}{2}\rho {\mathrm{C}}_{\mathrm{d}}\mathrm{A}}{1-\lambda }.$$

(20)

In a similar manner, using expression (11), the total ascent time can be discretized as

$$T=\underset{0}{\overset{T}{\int }}\mathrm{d}t=\sum _{j=1}^N\left(t_{j+1}-t_j\right)=\sum _{j=1}^N\frac{L_j}{V_{(j)}}.$$

(21)

Along with these considerations, using work (19) and time (21), the average power along the ascent is

$$\overline{P}=\frac{W}{T}.$$

(22)

Returning to model (2), its discretized form can be written as

$$P(t)=\left(\alpha \left(M_j+{\gamma }_j\right)+\beta V{(t)}^2\right)V(t),t_j\le t<t_{j+1},\mathrm{for}j=1,\dots ,N,$$

(23)

where ${\gamma }_j:=g\left({\mathrm{C}}_{\mathrm{rr}}\cos {\theta }_j+\sin {\theta }_j\right)$ . Since the jth acceleration, $M_j$ , and slope angle, ${\theta }_j$ , are discontinuous on subsequent segments, the power along the ascent is also discontinuous. Thus, the power at the endpoints of the jth segment is

$$P(t_j)=\left(\alpha \left(M_j+{\gamma }_j\right)+\beta V_j^2\right)V_j\mathrm{and}P(t_{j+1})=\left(\alpha \left(M_j+{\gamma }_j\right)+\beta V_{j+1}^2\right)V_{j+1}.$$

(24)

3. Minimum ascent time

3.1. Preliminary comments

According to Bos et al. [1–3], if the only constraint is a given average power, the optimal strategy for minimizing the ascent time of a cyclist is by a constant speed. To obtain this result, the authors neglected changes in kinetic energy and assumed that each discretized segment of the ascent is traversed at a constant speed with discontinuous speed between segments, since in the limit—as the number of segments tends to infinity—any speed can be approximated with arbitrary accuracy. The authors also considered the case of an average-power constraint subject to a maximum power, $P_{\max }$ , which is a scenario that might arise if the constant-speed solution requires power along portions of the ascent that exceeds a cyclist’s physical capacity. Hereafter, we refer to the constant-speed case as the ascent with a flying start. In that case, the optimal strategy is to fix the speed along those portions to correspond to $P_{\max }$ and calculate a different, albeit constant, speed with which the rest of the ascent is completed.

In this section, we revisit the minimum-ascent-time problem but include the power associated with changes in kinetic energy. We assume the speed to be piecewise linear and continuous between segments and the average power to be constrained to a specified average power, $P_0$ . Also, we consider the case of an average-power constraint subject to power along each segment bounded by a minimum, $P_{\min }$ , and maximum, $P_{\max }$ , power. In other words, the discretized problem consists of finding speeds $V_j\ge 0$ , $j=1,\dots ,N$ , that solve the problem

$$\begin{array}{ccc}\hfill \mathrm{minimize}& T=\sum _{j=1}^N\frac{L_j}{V_{(j)}},\hfill & \hfill \\ \hfill \mathrm{subject}\mathrm{to}& \overline{P}=P_0,\hfill \\ \hfill & P_{\min }\le P(t_j)\le P_{\max },\hfill \\ \hfill & P_{\min }\le P(t_{j+1})\le P_{\max }.\hfill \end{array}$$

(25)

For the numerical considerations, we use three constant-grade ascents of $\mathcal{G}=\{10\%,20\%,30\%\}$ such that each ascent has an elevation gain of $\mathrm{\Delta }Y=100\mathrm{m}$ . To achieve these grade percentages, the horizontal distance of each ascent is $\mathrm{\Delta }X=\{1000\mathrm{m},500\mathrm{m},333.3\mathrm{m}\}$ . Following Bos et al. [3, Section 4.1], the length of each ascent is $L=\sqrt{1+{\mathcal{G}}^2}\mathrm{\Delta }X=\{1004.9876\mathrm{m},509.9020\mathrm{m},348.0102\mathrm{m}\}$ . Also, for ease of comparison with Bos et al. [1–3], we let $m=70\mathrm{kg}$ , $g=9.81{\mathrm{m}∕s}^2$ , $\rho =1.2{\mathrm{kg}∕m}^3$ , ${\mathrm{C}}_{\mathrm{d}}\mathrm{A}=0.3{\mathrm{m}}^2$ , ${\mathrm{C}}_{\mathrm{rr}}=0.005$ , and $\lambda =0.02$ . Let us remark that the formulation presented in Section 2.2 is valid as long as the ascent is monotonically increasing (e.g., Bos et al. [3]). In particular, we use a constant-grade ascent for ease of illustration, not as a necessary simplification of the formulation.

3.2. Average-power constraint

Using MATLAB [10]’s fmincon optimization function, we seek to minimize the ascent time under the constraint of an average power, which we set to be $\overline{P}=300\mathrm{W}$ . As shown in Figure 1, the solution tends to a Heaviside step at the very beginning of the ascent, to increase the speed from zero to a constant speed of $3.9441\mathrm{m}∕s$ , which is maintained for the remainder of the ascent, except for another Heaviside step at the end of the ascent to return to zero. The resulting ascent time is $T=254.8223\mathrm{s}$ . As shown in Figure 1, due to sampling, the Heaviside step entails the Gibbs phenomenon.

Figure 1.

Segment speeds that minimize the ascent time along the 10%-grade ascent. The Heaviside step exhibits the Gibbs phenomenon due to a discretized formulation. The speed in the omitted region, marked by slashes on either side of the dotted line, is constant at $V=3.9441m∕s$ and corresponds to $P=300.0186W$ . Black dots correspond to $V_j$ and black lines to their interpolation.

The Gibbs phenomenon is well known in Fourier analysis. It arises if one approximates, for example, a square wave by a finite Fourier series. The Fourier series is smooth—indeed infinitely so—and hence, there is a conflict between the approximated square wave and the smooth approximating series. This results in over- and under-shoots known as “ringing.” The Gibbs analysis quantifies the percentage of overshoot. Our case, due to the finite-dimensional nature of the approximant, is analogous.

Remark 1. From Figure 1, it is evident that the optimal strategy is to accelerate as fast as the maximum-power constraint permits and thereafter maintain a constant speed, essentially to the end. The higher the maximum power permitted, the steeper and more concentrated is the initial acceleration. This discontinuous behavior creates a conflict between the true ideal optimal speed and what is permitted by the finite-dimensional piecewise linear continuous model. In such a case, in a manner similar to the standard Gibbs phenomenon, the optimal numerical solution exhibits compensatory “ringing,” which has to be kept in mind, and which we address in the next section.

Both the constant speed and the ascent time are consistent with the results found by Bos et al. [1–3]—achieved with the same discretization and average-power constraint, but with a flying start—who found the constant speed of $3.9439\mathrm{m}∕s$ and the ascent time of $254.8206s$ . However, an instantaneous acceleration is physically unreasonable; it implies an infinite power.

For N = $10000$ and the $10\%$ -grade ascent, the Gibbs phenomenon results—at the end of the first and beginning of the last segment—in power approaching $60000\mathrm{W}$ and $-60000\mathrm{W}$ , respectively. We summarize these results, along with those for the $20\%$ - and $30\%$ -grade ascents, in Table 1, which, given the fine discretization of N = $10000$ , are nearly identical to those of the flying start, found by Bos et al. [1–3].

Table 1.

Ascent times, T, and minimizing constant speed, V, for three ascent profiles of increasing grade percentage, $\mathcal{G}$ , under the constraint of constant-average power.

				Flying start		Standing start
$\mathcal{G}[\%]$	$\mathrm{\Delta }X[\mathrm{m}]$	$\mathrm{\Delta }Y[\mathrm{m}]$	$L[\mathrm{m}]$	$V[\mathrm{m}∕\mathrm{s}]$	$T[\mathrm{s}]$	$V[\mathrm{m}∕\mathrm{s}]$	$T[\mathrm{s}]$
10	1000	100	$1004.9876$	3.9439	$254.8206$	3.9441	$254.8223$
20	500	100	$509.9020$	2.1174	$240.8104$	2.1176	$240.8107$
30	$333\frac{1}{3}$	100	$348.0102$	1.4627	$237.9202$	1.4630	$237.9204$

3.3. Average-power constraint subject to bounded power

To render the solution physically reasonable, we constrain the power on each segment by its minimum and maximum value. To do so, we use expression (24), at the endpoints of jth segment, so that

$$P_{\min }\le P_j\le P_{\max }\mathrm{and}{P}_{\min }\le P_{j+1}\le P_{\max },\mathrm{for}j=1,\dots ,N.$$

(26)

Returning to MATLAB [10]’s fmincon optimization function, we seek to minimize the ascent time under the constraint of average power, $\overline{P}=300\mathrm{W}$ , together with inequalities (26), with $P_{\min }=0\mathrm{W}$ and $P_{\max }=450\mathrm{W}$ . Herein, $P=0$ corresponds to the cyclist not expending any effort, whereas negative power, $P<0$ , would correspond to the cyclist increasing their internal energy, which is physically unreasonable.

Choosing the $10\%$ -grade ascent and setting N = $10000$ , we find that the minimum ascent time is T = $254.9527\mathrm{s}$ , obtained with a constant speed of V = $3.9767\mathrm{m}∕s$ . At the beginning of the ascent, it requires $2.5986\mathrm{s}$ to increase the speed from a standstill to the constant speed without exceeding $P_{\max }$ . At the end of the ascent, it requires $3.3258\mathrm{s}$ to decrease the speed from the constant speed to a final speed of $V_{N+1}$ = $0.5175\mathrm{m}∕s$ without reaching negative powers. The segment speed and power are shown in Figure 2.

Figure 2.

Segment speed and power that yield a minimum ascent time of $T=254.9527$ s along the 10%-grade ascent, under the constraint of a constant-average power, $\overline{P}=300$ W and subject to power bounded by $P_{\min }$ and $P_{\max }$ . The speed in the omitted region, marked by slashes on either side of the dotted line, is constant at $V=3.9811m∕s$ and corresponds to $P=302.6858W$ . Black dots correspond to $V_j$ in (a), and to $P_j$ and $P_{j+1}$ in (b); black lines correspond to the interpolation of each data point’s respective endpoints.

These results, along with those for the $20\%$ - and $30\%$ -grade ascents, are summarized in Table 2.

Table 2.

Ascent times, T, and minimizing constant speed, V, for three ascent profiles of increasing grade percentage, $\mathcal{G}$ , under the constraint of constant-average power and subject to power bounded by $P_{\min }$ and $P_{\max }$ . Initial speed of the launched efforts is $V_1=11.7217\mathrm{m}∕s$ , which corresponds to $P=300\mathrm{W}$ on a flat section.

	Flying start		Standing start		Launched start
$\mathcal{G}[\%]$	$V[\mathrm{m}∕\mathrm{s}]$	$T[\mathrm{s}]$	$V[\mathrm{m}∕\mathrm{s}]$	$T[\mathrm{s}]$	$V[\mathrm{m}∕\mathrm{s}]$	$T[\mathrm{s}]$
10	3.9441	$254.8223$	3.9811	$254.9527$	4.0977	$240.9870$
20	2.1176	$240.8107$	2.1250	$240.8172$	2.1707	$225.7489$
30	1.4630	$237.9204$	1.4652	$237.9213$	1.4888	$222.4865$

The deceleration at the end of the ascent found in the numerical solution and shown in Figure 1 may appear surprising, since it results in a longer ascent time than maintaining a constant speed would. However, it is intuitively understandable. Since the solution minimizes the ascent time T, the average-power constraint in equation (22) implies that the total work W is also minimized. The numerical solution accomplishes this objective by minimizing $\mathrm{\Delta }K$ , which, since herein the initial speed is taken to be zero, implies a final speed that approaches zero. In physical terms, maintaining a substantial speed at the end of the ascent would result in greater than minimum work being performed.

3.4. Launched start

Let us consider a case for which the cyclist approaches the ascent along a horizontal road. Given the power along that road that is equal to the average-power constraint, $\overline{P}$ , along the hill, the speed at the beginning of the ascent is greater than the subsequent ascent speed. Hence, we refer to such a scenario as a launched start. Notably, such situations are common in competitions, where the last part of the timetrial is an ascent to the finish line.

Letting P = $300W$ and θ = 0 in expression (2), we obtain, through algebraic manipulation, the depressed cubic equation

$$V^3+\frac{\alpha {\mathrm{C}}_{\mathrm{rr}}}{\beta }V-\frac{P}{\beta }=0,$$

(27)

herein, the solution is V= $11.7217\mathrm{m}∕s$ . Using this value as the initial uphill speed, with $\mathcal{G}$ = $10\%$ , $\overline{P}$ = $300W$ , $P_{\min }$ = $0\mathrm{W}$ , and $P_{\max }$ = $450\mathrm{W}$ , the minimum ascent time is T = $240.9870\mathrm{s}$ . As illustrated in Figure 3, the solution tends to a constant speed of $4.0977\mathrm{m}∕s$ , which corresponds to a power of $312.6285\mathrm{W}$ , remains at that speed for the majority of the ascent and decreases to a final speed of $V_{N+1}$ = $0.6185\mathrm{m}∕s$ . The initial decrease in speed takes $6.4175\mathrm{s}$ , whereas the final decrease takes $3.3408\mathrm{s}$ . These results, along with those for the $20\%$ - and $30\%$ -grade ascents, are summarized in “Launched start” column of Table 2. Note that, since $V_1\ne 0$ , the first term in parentheses in expression (5) is $\frac{1}{2}m\left(V_T^2-V_1^2\right)$ , which consequently modifies the subsequent discretization.

Figure 3.

Segment speeds that yield a minimum ascent time of $T=240.9870\mathrm{s}$ in a launched effort along the 10%-grade ascent, under the constraint of a constant-average power, $\overline{P}=300$ W and subject to power bounded by $P_{\min }$ and $P_{\max }$ . The speed in the omitted region, marked by slashes on either side of the dotted line, is constant at $V=4.0977m∕s$ and corresponds to $P=312.6285W$ .

One of the main applications of a launched-start model is to gain an insight into timetrials composed of flat and steep sections by examining a strategy of stopping before the ascent to change a bicycle that optimizes a flat course, where the effort is dominated by overcoming air resistance, to a bicycle for an ascent, where the effort is dominated by overcoming gravity. Comparing the ascent times in Table 2 for the standing and launched starts, we notice a significant time loss due to stopping and restarting. One needs to decide whether or not such a loss is more than compensated by a choice of a more appropriate bicycle.

Let us consider a flat course ending with a $1004.9876m$ ascent of 10%. For the flat portion of the race, the cyclist uses a bicycle that results in m = $70\mathrm{kg}$ ; the bicycle has disk wheels and handlebar extensions to minimize air resistance. Hence, we set the resulting air-resistance coefficient to ${\mathrm{C}}_{\mathrm{d}}\mathrm{A}$ = $0.2{\mathrm{m}}^2$ . Letting P = $300\mathrm{W}$ , θ = 0, g = $9.81{\mathrm{m}∕s}^2$ , ρ = $1.2{\mathrm{kg}∕m}^3$ , ${\mathrm{C}}_{\mathrm{rr}}$ = 0.005, and λ = 0.02, in equation (2), we obtain $V=13.4089\mathrm{m}∕s$ .

If there is no bicycle change, the cyclist starts the ascent with $V_1$ = $13.4089\mathrm{m}∕s$ . Since the position for an ascent is less aerodynamic, we set ${\mathrm{C}}_{\mathrm{d}}\mathrm{A}$ = $0.3{\mathrm{m}}^2$ . The resulting ascent time is T = $237.3167\mathrm{s}$ .

If the cyclist changes for a lighter bicycle that results in m = $68\mathrm{kg}$ , there is a need to restart from $V_1=0m∕s$ . Again, for a less aerodynamic climbing position, we set ${\mathrm{C}}_{\mathrm{d}}\mathrm{A}$ = $0.3{\mathrm{m}}^2$ . The resulting ascent time is $T=248.4836\mathrm{s}$ .

Comparing these ascent times, we see that, in this case as well as for the two other cases shown in Table 3, restarting with a lighter bicycle is not sufficiently beneficial; launched starts still result in shorter times. Nevertheless, comparing the ascent time for standing starts in Table 2 with corresponding times in Table 3, we notice a significant, albeit overall insufficient, improvement with a lighter bicycle.

Table 3.

Ascent times, T, and minimizing constant speed, V, for three ascent profiles of increasing grade percentage, $\mathcal{G}$ , under the constraint of constant-average power and subject to power bounded by $P_{\min }$ and $P_{\max }$ . Mass of the standing start efforts is $m=68\mathrm{kg}$ . Initial speed of the launched efforts is $V_1=13.4089\mathrm{m}∕s$ , which corresponds to $P=300\mathrm{W}$ on a flat section with ${\mathrm{C}}_{\mathrm{d}}\mathrm{A}=0.2{\mathrm{m}}^2$ .

	Standing start		Launched start
	(m = $68\mathrm{kg}$ )		( $V_1$ = $13.4089\mathrm{m}∕s$ )
$\mathcal{G}[\%]$	$V[\mathrm{m}∕\mathrm{s}]$	$T[\mathrm{s}]$	$V[\mathrm{m}∕\mathrm{s}]$	$T[\mathrm{s}]$
10	4.0920	$248.4836$	4.1246	$237.3167$
20	2.1884	$234.0619$	2.1787	$221.5468$
30	1.5094	$231.1644$	1.4929	$218.0571$

The formulation presented herein allows us to adjust input parameters for a problem at hand to help in choosing a race strategy by quantifying the wisdom of an adage that the longer and steeper the ascent the more likely one benefits from a bike change, as illustrated in a specific example by Bos et al. [11, Appendix A]. To arrive at a final decision, a competitor also needs to take into account time spent for such actions as slowing down at the end of a flat section and the bike change itself, which are not included in this formulation.

4. Conclusion

Generalizing the formulation of Bos et al. [1–3] by not restricting the initial speed to a value determined by the model and its optimization, we gain an insight into the optimization process and its results. Similarly to the result proven analytically by Bos et al. [3] in the aforementioned formulation, the minimization tends to a constant speed for the ascent regardless of the initial speed and the ascent profile. Notably, as stated in Tables 1 and 2, the ascent times are almost identical to those in Bos et al. [3].

In the present paper, for conciseness and ease of comparison, we use only constant-grade ascents. However, we have confirmed by numerical experimentation that the minimization tends to a constant speed for arbitrary ascent profiles, such as the ones discussed by Bos et al. [3], and shown in their Figures 4(b)–(e). This remains true for models subject to the average-power, minimum-power, and maximum-power constraints. If the two latter constraints are included, a resulting constant speed is different from the one under the average-power constraint alone to accommodate the parts of the ascent that are restricted by the power bounds.

This generalization also renders the model more flexible and empirically adequate. As discussed in Section 3, the model is ready to be used for uphill timetrials and for timetrials composed of flat and steep sections. Future work might focus on presenting strategies for specific races and examining their results.