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Abstract

VAM (velocità ascensionale media) is a measurement that quantifies a cyclist’s climbing ability. We show that to minimize the time to attain a given height gain—which is tantamount to maximizing VAM—a cyclist should climb as steep a constant-grade hill as possible. Apart from the power-to-weight ratio, the limit of steepness is imposed by such factors as the efficiency of pedalling, which is related to feasible cadence, maintaining balance, preventing lifting of the front, and skidding of the rear, wheel. This article is focused on consequences of the power available to the cyclist, which can be viewed as a necessary condition to examine other aspects of climbing strategy. We show that—for given start and end points, and for any fixed average-power constraint—the brachistochrone, which is the trajectory of minimum ascent time, is the straight line connecting these points, covered with a constant speed, which along such a line is equivalent to a constant power. This is in contrast to the classical solution of a descent brachistochrone under gravity, which is a cycloid along which the speed is not constant.

Keywords

VAM average power power-to-weight ratio height-gain maximization brachistochrone phenomenological model

1. Introduction

VAM (velocità ascensionale media) is the average ascent velocity. It is a measurement of the rate of ascent that quantifies a cyclist’s climbing ability.

In this article, we solve an optimization problem motivated by and applied to cycling. Specifically, we find the hill-climbing trajectory that minimizes the ascent time. While this problem is not directly concerned with the mechanics of continuous media, there are interesting parallels between it and the problem of wave propagation in layered media. In particular, in the geometrical approximation to wave propagation, the trajectory of a ray can be found from Fermat’s principle. This is a variational problem whose solution, the ray trajectory, extremizes—typically minimizes—the travel time, in a close mathematical analogy to the problem considered here.

This article is a direct follow-up of Bos et al. [1]; therein, the authors reach the same conclusions but with an argument where the imposed assumptions overly reduce the candidate ascents; hence, this follow-up. Ascent optimizations are also examined by Bos et al. [2 –5], and the results of these four articles are used herein. There is, however, an important distinction between Bos et al. [1] and Bos et al. [2 –5]. In the 2024 and 2025 articles, we optimize the ascent strategy for a given uphill; in other words, the ascent profile is known. In the 2021 article—as well as in the present article—we seek the profile to maximize the ascent rate for a given average power. In other words, we seek the corresponding brachistochrone. Notably, brachistochrone problems in relation to cycling are discussed also in other contexts [e.g., 6].

We begin this article by examining the expression relating power and speed as a function of steepness. We prove two lemmas that lead us to conclude that to minimize the time to attain a given height gain, a cyclist should ride as steep a climb as possible. We conclude the article with numerical insights and a brief discussion, where we comment on certain qualifiers of this conclusion.

2. Formulation

Following our previous work [2 –5], we let

\begin{matrix} P = \frac{\overset{change in speed}{\overset{︷}{m \frac{dV}{dt}}} + \overset{change in elevation}{\overset{︷}{m g \sin θ}} + \overset{rolling resistance}{\overset{︷}{C_{rr} \underset{normal force}{\underset{⏟}{m g \cos θ}}}} + \overset{air resistance}{\overset{︷}{\frac{1}{2} C_{d} A ρ V^{2}}}}{\underset{drivetrain efficiency}{\underset{⏟}{1 - λ}}} V, \end{matrix}

(1)

where P is the power, V is the ground speed, m is the mass of the bicycle-cyclist system, g is the acceleration due to gravity, θ is the slope of the hill, $C_{rr}$ is the rolling-resistance coefficient, $C_{d} A$ is the air-resistance coefficient, ρ is the air density, and λ is the drivetrain-resistance coefficient.

Bos et al. [3, Section 3] prove that, given an average-power constraint, the ascent time is minimized if a cyclist maintains a constant ground speed, regardless of the slope. In particular, if the average power, $\bar{P}$ , is constrained to a given value, $P_{0}$ , over the ascent time, T, namely,

\bar{P} = \frac{1}{T} \int_{0}^{T} P dt = P_{0},

then the optimal constant ground speed, V, can be found by solving

\begin{matrix} m g V \frac{H + C_{rr} D}{L} + \frac{1}{2} C_{d} A ρ V^{3} = (1 - λ) P_{0}, \end{matrix}

(2)

where H is the height gain, D is the horizontal distance, and L is the arclength of the ascent curve [3, Appendix G].

The work done in reaching any speed—before starting the ascent—is not included in the average power for the ascent. In any case, such an effect is negligible for longer ascents, which are required to quantify the cyclist’s climbing ability using VAM.

Remark 1. In accordance with equation (2), for fixed H and D, the optimal constant value of V is a function of L alone.

The ascent time is

\begin{matrix} T = \frac{L}{V} . \end{matrix}

(3)

Given H and D, the optimal speed, $V = V (L)$ , is defined implicitly by equation (2), as a function of L; hence, equation (3) expresses implicitly T as a function of L.

Let us state the following proposition, whose proof follows from Lemma 1.

Proposition 1. For given start and end points, the curve of the minimum ascent time—for any fixed average-power constraint—is the straight line connecting these points.

Lemma 1. For $L > 0$ ,

\frac{dT}{dL} > 0 .

Proof. Since

\frac{dT}{dL} = \frac{V - L V^{'} (L)}{V^{2}},

where $V^{'} (L) : = dV ∕ dL$ , we must show that

V - L V^{'} (L) > 0, for L > 0 .

Let us write equation (2) as

a \frac{V}{L} + b V^{3} = c,

where

a : = m g (H + C_{rr} D) > 0, b : = \frac{1}{2} C_{d} A ρ > 0, c : = (1 - λ) P_{0} > 0 .

Differentiating implicitly with respect to L, we obtain

\begin{matrix} a (\frac{L V^{'} - V}{L^{2}}) + 3 b V^{2} V^{'} & = 0 \\ ⟹ a (L V^{'} - V) + 3 b L^{2} V^{2} V^{'} & = 0 \\ ⟹ V^{'} (a L + 3 b L^{2} V^{2}) - a V & = 0, \end{matrix}

and, hence,

V^{'} (L) = \frac{a V}{a L + 3 b L^{2} V^{2}} > 0 .

Substituting this expression into $V - L V^{'} (L)$ , we obtain

\begin{matrix} V - L V^{'} (L) & = V - \frac{a V}{a + 3 b L V^{2}} \\ = \frac{V (a + 3 b L V^{2}) - a V}{a + 3 b L V^{2}} \\ = \frac{3 b L V^{3}}{a + 3 b L V^{2}} > 0, \end{matrix}

as claimed. □

In other words, the time of ascent is an increasing function of the length of the ascent path. Since $dT ∕ dL$ is strictly greater than zero, T is a strictly increasing function of L and, therefore, the value of L that minimizes T is unique. Hence, it follows that the minimum time is attained for the curve of minimum length, which is the straight line connecting the start and end points of the ascent, as stated in Proposition 1.

Remark 2. The condition $dT ∕ dL > 0$ says, in words, that the longer the length of the ascent, the greater the time required, and so the most direct ascent—a straight line—also requires the least amount of time. This may appear, at first sight, to be obvious. However, even for motion in a horizontal plane, a straight line might not result in a shortest time, as would be the case with changing rolling friction due to variable road conditions, which is akin to ray bending in accordance with Fermat’s principle in layered media. For motion in a vertical plane, the solution of the classical brachistochrone problem is a cycloid and not a straight line. These examples suggest that, for an ascent, there are subtleties that need to be explored and thus that Lemma 1 is a nontrivial result.

Remark 3. Following expressions proven in Lemma 1, it follows that

\frac{dT}{dL} = \frac{V - L V^{'} (L)}{V^{2}} = \frac{3 b L V}{a + 3 b L V^{2}} \frac{a L}{a L} = \frac{3 b L^{2}}{a} \frac{a V}{a L + 3 b L^{2} V^{2}} = \frac{3 b L^{2}}{a} \frac{dV}{dL},

which is an interesting relation.

We now consider the question of finding, among straight lines with a given height gain, H, the one that gives the shortest ascent time. The answer, as we show, is—in the limit—a vertical climb.

Lemma 2. For $L = \sqrt{H^{2} + D^{2}}$ , with a given height gain, H, and a varying horizontal distance, $D > 0$ ,

\frac{dT}{dD} > 0 .

Proof. Given H and $L = \sqrt{H^{2} + D^{2}}$ , equation (2) implicitly states V as a function of D. Differentiating equation (3) with respect to D, we obtain

\begin{matrix} \frac{dT}{dD} = \frac{D V - L^{2} V^{'}}{L V^{2}}, \end{matrix}

(4)

where $V^{'} : = dV ∕ dD$ , and we use the fact that $dL ∕ dD = D ∕ L$ . Letting

a (D) : = m g (H + C_{rr} D), b : = \frac{1}{2} C_{d} A ρ > 0, c : = (1 - λ) P_{0},

we write equation (2) as

a (D) \frac{V}{L} + b V^{3} = c .

Differentiating implicitly with respect to D, we obtain

\begin{matrix} a^{'} (D) \frac{V}{L} + a (D) (\frac{V^{'} (D)}{L} - \frac{V}{L^{2}} L^{'} (D)) + 3 b V^{2} V^{'} (D) \\ = a^{'} (D) \frac{V}{L} + a (D) (\frac{V^{'} (D)}{L} - \frac{V}{L^{2}} \frac{D}{L}) + 3 b V^{2} V^{'} (D) \\ = V^{'} (D) (\frac{a (D)}{L} + 3 b V^{2}) + \frac{a^{'} (D) V}{L} - \frac{a (D) V D}{L^{3}} = 0; \end{matrix}

hence,

V^{'} (D) (\frac{a (D) + 3 b V^{2} L}{L}) = \frac{V}{L^{3}} (- a^{'} (D) L^{2} + a (D) D) .

But

\begin{matrix} - a^{'} (D) L^{2} + a (D) D & = - m g C_{rr} (H^{2} + D^{2}) + m g (H + C_{rr} D) D = - m g C_{rr} H^{2} + m g H D \\ = m g H (D - C_{rr} H) . \end{matrix}

Consequently,

V^{'} (D) = \frac{V}{L^{2}} \frac{m g H (D - C_{rr} H)}{a (D) + 3 b V^{2} L},

so that

\begin{matrix} DV - L^{2} V^{'} (D) & = DV - V \frac{m g H (D - C_{rr} H)}{a (D) + 3 b V^{2} L} \\ = \frac{V}{a (D) + 3 b V^{2} L} (D (a (D) + 3 b V^{2} L) - m g H (D - C_{rr} H)) \\ = \frac{V}{a (D) + 3 b V^{2} L} (D (m g (H + C_{rr} D)) + 3 b V^{2} LD - m g HD + m g C_{rr} H^{2}) \\ = \frac{V}{a (D) + 3 b V^{2} L} (m g C_{rr} (D^{2} + H^{2}) + 3 b V^{2} LD) \\ = \frac{V}{a (D) + 3 b V^{2} L} (m g C_{rr} L^{2} + 3 b V^{2} LD) > 0 \end{matrix}

and thus—by equation (4)— $dT ∕ dD > 0$ . □

Consequently, a cyclist wishing to minimize the time to attain a given height gain should ride as steep a constant-grade climb as possible.

3. Numerical insights

3.1. Speed, its vertical component and ascent time as functions of slope

In this section, we present numerical insights into the ground speed, V, its vertical component and the ascent time, as a function of the slope, θ. To do so, we return to model (1) with $P = 300 W$ , $m = 68 kg$ , $g = 9.81 {m ∕ s}^{2}$ , $ρ = 1.2 {kg ∕ m}^{3}$ , $C_{d} A = 0.3 m^{2}$ , $C_{rr} = 0.005$ and $λ = 0.02$ , and we let $θ \in (5^{\circ}, 2 0^{\circ})$ , which corresponds to slopes in the range $(8.7489 %, 36.3970 %)$ .¹ To solve for speed, we use the fact that expression (1) becomes a cubic equation in V, with only one positive real solution [3, Section 4.2].

The ground speed is shown in Figure 1. As expected—given the power constraint—it decreases with the increase of steepness.

Figure 1.

$V (θ)$ : optimal constant ground speed as a function of slope.

The vertical component of that speed is $V (θ) \sin θ$ , which is shown in Figure 2. In accordance with the results in Section 2, it increases with the increase in steepness.

Figure 2.

$V (θ) \sin θ$ : vertical speed as a function of slope.

The ascent time for the elevation gain of $1000 m$ is shown in Figure 3. Again, in accordance with Section 2, it decreases with steepness.

Figure 3.

$T (θ)$ : ascent time for the elevation gain of $1000 m$ as a function of slope.

3.2. Relations between power, mass and optimal slopes

In this section, we present numerical insights into relations between power, mass and optimal slopes. Figures 1 –3 examine ascents under the fixed power constraint, $P = 300 W$ , illustrating, respectively, ground speed, vertical speed and ascent time as a function of slope. Figure 4 examines ascents under the constraint of constant ground speed, $V = 1.5 m ∕ s$ . For a given mass, m, the intersection of a vertical line with a contour line representing a slope corresponds to the power, P, required to achieve the ascent along that slope with $V = 1.5 m ∕ s$ . As expected, an increase in mass or steepness requires greater power to maintain the constant-speed ascent.

Figure 4.

Contour lines showing maximum slope climbable with $V = 1.5 m ∕ s$ , for a given mass, m, and power, P.

We assume $V = 1.5 m ∕ s$ to be the lower bound for efficiency of pedalling in uphill cycling. For available gear ratios, it corresponds to a cadence of about 50 RPM. Lower speeds would result in balance issues and lower cadences in an exceeding high pedal torque for each revolution. VAM as a function of cadence, gear ratio and force is discussed by Bos et al. [1, Section 3.3].

We also need to consider pragmatic limitations in handling a bicycle, such as lifting of the front wheel and skidding of the rear wheel due to steepness of a climb. From empirical studies, we infer that the steepest ascents achievable in cycling are about $30 %$ ; hence, the values in the upper left-hand corner of Figure 4 are unrealistic. To gain an insight into consequences of this limitation, let us consider Figure 5, where we consider a common metric in competitive cycling, namely, the power-to-weight ratio. Riders whose power-to-weight ratio—for a given time interval—is greater than about $5 W ∕ kg$ should not strive to maximize their VAM by choosing a steeper uphill but by remaining at about $30 %$ and increasing the ground speed, V.

Figure 5.

Slope as a function of power-to-weight ratio for $V = 1.5 m ∕ s$ .

For instance, letting $P = 600 W$ and $m = 75 kg$ , which corresponds to $P ∕ m = 8 W ∕ kg$ , and considering the slope of $30 %$ , as well as $ρ = 1.2 kg ∕ m^{3}$ , $CdA = 0.3 m^{2}$ , $Crr = 0.005$ and $λ = 0.02$ , we find—according to model (1)—the ground speed $V = 2.7188 m ∕ s$ . Hence, for such a rider, the optimal strategy to maximize VAM is the slope of $30 %$ and $V = 2.7188 m ∕ s$ , not the slope of $62 %$ and $V = 1.5 m ∕ s$ , even though, as expected in view of Lemmas 1 and 2—to gain $1000 m$ — $T (30 %) = 1280 s > T (62 %) = 1262 s$ ; the latter, however, is unrealistic.

4. Discussion and conclusion

As stated by Lemmas 1 and 2, respectively, the optimal ascent to maximize the VAM—for a given average power—is a constant-grade hill with the highest grade. As proven by Bos et al. [3, Section 3]—for a given average power—the ascent time is minimized by a constant ground speed, regardless of slope’s steepness and shape. Hence, constancy of grade and speed imply a constancy of instantaneous power.

In other words, for given start and end points as well as a fixed average-power constraint, the brachistochrone is the straight line connecting these points. VAM is maximized by the shortest distance covered with a constant speed corresponding to that average power, which is also the instantaneous power. This is in contrast to the classical solution of a descent brachistochrone under gravity, which is a cycloid along which the speed is not constant. Herein, the shortness of the straight line, which is tantamount to its steepness, is subject to the power-to-weight ratio and to pragmatic issues of pedalling efficiency and bicycle handling.

Footnotes

Acknowledgements

We wish to acknowledge insightful comments of Scott Anderson, Mikhail Kochetov, Andrea Oliveri and an anonymous reviewer, as well as proofreading of David Dalton.

Authors’ note

There is no supplementary material. Interested readers might contact the corresponding author for computations with a variety of input parameters.

ORCID iD

Theodore Stanoev

Funding

The authors disclosed receipt of the following financial support for the research, authorship and/or publication of this article: The research is partially supported by the NSERC Discovery Grant RGPIN-2018-05158 of M.A.S.

Declaration of conflicting interests

The authors declared no potential conflicts of interest with respect to the research authorship and/or publication of this article.

Notes

References

Bos

Slawinski

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, et al. On minimizing cyclists’ ascent times: part I. arXiv 2025. DOI: 10.48550/arXiv.2403.03363.

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, et al. On minimizing cyclists’ ascent times: part II. arXiv 2025. DOI: 10.48550/arXiv.2503.03235.

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On the brachistochrone problem for cycling ascents

Abstract

Keywords

1. Introduction

2. Formulation

3. Numerical insights

3.1. Speed, its vertical component and ascent time as functions of slope

3.2. Relations between power, mass and optimal slopes

4. Discussion and conclusion

Footnotes

Acknowledgements

Authors’ note

ORCID iD

Funding

Declaration of conflicting interests

Notes

References