On level terrain a bicyclist is faster than a runner.
On a steep slope a runner is faster than a bicyclist.
(That's why you see bicyclists pushing their bikes uphill.)

Can we calculate the angle where it becomes more advantageous to discard the bicycle and start running? If so, what is the formula?


Apparently we can calculate the steepest gradient for a bicycle. Rhett Allain (professor of physics at Southeastern Louisiana University) calculates a "a maximum incline of 38.7°" (80%) on a dry road. Cyclist Magazine writes, 60% is probably more realistic.

So the angle where cycling becomes less efficient than running must be at least a little less than that.

Let's assume the most basic model with "ideal" circumstances: a straight road with a constant incline, dry, asphalt, no wind, etc.

(As the question is probably already difficult enough, please don't artificially complicate it by introducing exceptional circumstances like nighttime, rain, hail, oncoming traffic, a flat tyre, alien attacks, etc.)

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    Jan 21, 2022 at 15:57

4 Answers 4


We can at least calculate the angle at which the bike can no longer make progress. In order to maintain any nonnegative vertical velocity, the distance-averaged output force applied by the cyclist to the ground via the rear wheel must equal the force of gravity down an inclined plane:

$\bar F_\text{out} = g(m_\text{man}+m_\text{bike})\sin(\theta)$

Let $D$ be twice the crank length of the bicycle pedal, that is, twice the distance from the pedal to the center of the front gear.

Let $L$ be the minimum displacement of the wheel along the slope per half revolution of the crank

$L = \frac{\pi}{2} \times \text {(Wheel diameter) (number of front teeth) / (number of rear teeth)}$

The ratio of output to input force for the bicycle is $R = \frac {\pi D}{2L} = F_\text{out}/F_\text{in}$

So at the point where $\bar F_\text{in} = \frac{g}{R}(m_\text{man}+m_\text{bike})\sin(\theta)$ exceeds the maximum force the athlete can supply to the pedal without falling off (or leaping off), the cyclist can no longer make progress. This must be less than $gm_\text{man}$, or the cyclist will jump off.$^1$

$\theta_\text{max} \lt \arcsin\left(\frac{Rm_\text{man}}{m_\text{man}+m_\text{bike}}\right)$

For a 70 kg man and a 10 kg bike, with a minimum $L$ of $25\pi$ cm and a $D$ of $34$ cm giving $R = 0.68$, that gives

$\theta_\text{max} \lt 37 ^\circ$

This is considerably higher slope than I'd consider possible to bicycle up, which probably reflects the fact that balancing your entire weight on one moving pedal of a slow-moving bicycle, just to keep from rolling down the hill, is much more difficult than just getting off and walking. (Not to mention slower, and likely to damage the chain or crank.)

Given Superman and a specially designed indestructible bike, any angle could be achieved, since the athlete could, hypothetically, pull up on the bike to counteract pushing down on the pedals and greatly exceed his own weight in thrust. However, such an athlete would be better suited to leaping the hill in a single bound.

1: I have assumed that the cyclist (except for Superman and his superbike) has no good way of pulling up on the bike in a normal cycling posture. It may be possible to exert more force on the crank by pulling up with the opposite foot, if the athlete is clipped in to the pedals. This would switch the problem from a question about maximum force to a question about the maximum lateral torque that can be applied without flipping the bike, which in turn is a question about how well and how far the athlete can alternate leaning back and forth to counterbalance. I think that puts the question out of reach of a first-principles approach.

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    $\begingroup$ Hm. you are on to something. I would consider also the torque. At some angle the applied torque will flip the bike, not spin the wheel. $\endgroup$
    – Stian
    Jan 13, 2022 at 8:55
  • $\begingroup$ Best answer so far, thank you! Your calculation appears pretty accurate, given that Rhett Allain (professor of physics at Southeastern Louisiana University) has calculates a similar value for the maximum incline: 38.7° (see link in question). Now if you could continue your calculations to compare the cyclist with the runner, that would make the answer perfect! $\endgroup$
    – summerrain
    Jan 13, 2022 at 12:08
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    $\begingroup$ Superman's bicycle still has a hard limit. The maximum angle in that case is limited by the location of the centre of mass and the contact point, as soon as the vertical from centre of mass falls behind the contact point the cyclist can no longer make progress. $\endgroup$
    – Separatrix
    Jan 13, 2022 at 15:26
  • $\begingroup$ Does this make the implicit assumption that the cyclist cannot lift up on the pedals (i.e. is not clipped in). Intuitively, it seems that a clipped cyclist could generate more force than $m_{man}g$ if they can pair an upward force on one pedal with the downward force on another. Cyclists do indeed do this, although its possible my intuition falls short in this extreme case. $\endgroup$
    – Cort Ammon
    Jan 13, 2022 at 16:21
  • $\begingroup$ @summerrain It wouldn't be an exacting comparison, but its worth looking at stairs. Humans can clearly move up stairs. Stairs are typically in the 30-40 degree range, but a google search shows that industrial applications have stairs that are in the 40-70 degree range. There's no particular limit to this with the physiology of the human body, although at some point we do have to start making toe-holds in the stairs to ensure that we can generate enough force to oppose gravity without slipping. $\endgroup$
    – Cort Ammon
    Jan 13, 2022 at 16:24

A typical racing bicycle tire has a circumference of just over 2 meters. A really low gear has a ratio of 1, meaning that one circle of the pedals turns the rear wheel just 1 time. So this would mean that pushing your right foot from 12 o'clock down to 6 o'clock and lifting it back up to 12 o'clock position would move the bike 2 meters forwards. So that would be the equivalent of taking two complete strides in running.

If running up a steep hill, I don't think you could achieve 1 meter forward for one complete stride. The reason is that with steepness, your stride become shorter. On flat land you could easily have strides longer than 1 meter, but not up a hill.

The thing about racing bicycles, with proper pedals, is that you have power on both the down stroke and the up stroke. In fact a good racer has power for the complete 360 degrees of the circle. This was a very important part of my training back when I was racing. But a runner only has power on the downstroke; the upstroke is completely wasted time and effort.

So if we assume there is no slippage with the tires on the ground, I can't see any point where it would be faster to run. However, I do know that in dirt bike racing, slippage in the mud is a major issue, so they often dismount and run (or more likely walk) up the hill.

So from a purely physics point of view, I think cycling would always be faster.

However, I think that if you were looking at flat land racing, and if the race was very short, say 10 meters, then running would probably be faster than cycling because the acceleration would be much slower for the cyclist while the runner can explode to over 1 meter per stride very quickly.

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    $\begingroup$ On a vertical cliff face, a climber (runner? well, some are really fast) would be faster than a bicycle. $\endgroup$
    – Jon Custer
    Jan 12, 2022 at 23:44
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    $\begingroup$ @ "So from a purely physics point of view, I think cycling would always be faster.": I don't think this is correct. See this question: "Why is it so much harder to ride a bike up a hill than to push it?" $\endgroup$
    – summerrain
    Jan 13, 2022 at 7:17
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    $\begingroup$ @JonCuster I think this question assumes perfect grip. So in the case of perfect verticality, it's more of a ladder versus a pedal-driven lift type deal. $\endgroup$
    – Arthur
    Jan 13, 2022 at 9:41
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    $\begingroup$ As someone who used to mountain bike a lot, the bike gets to be comparable in speed to walking around about the same point it starts to become impossible. Even if you ignore wheel spinning there's the phenomenon of being unable to pedal without wheelying. A good answer could at least address that! $\endgroup$ Jan 13, 2022 at 9:52
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    $\begingroup$ I should note that some bicycles go lower than 1:1, especially among cargo-oriented vehicles that need the extra torque and aren't subject to wheelie-ing. For instance, I own a large Dutch cargo hauler with a (fixed) gear ratio of 5:6. $\endgroup$ Jan 13, 2022 at 13:57

To avoid misunderstandings: Even though this post ends with a number I am only trying to establish a baseline on the back of an envelope. All concrete numbers are guesses or convenient numbers (hey, 0.1 m/s!), the concrete bio-mechanic assumptions are laughable etc., but I am convinced that I outlined the essence of the problem in case anybody wants to flesh it out with proper data.

To get an angle at the question we should first state the obvious: The speed (in a sustained equilibrium, without inertia considerations) cannot exceed the state when the maximum power (energy/time) the running or cycling person generates equals the gain in potential energy plus the friction losses (internally, in the muscles etc., and externally through air resistance and friction of bicycle bearings and ground contact). Higher speeds always increase friction and the rate at which potential energy grows; at some point there is no power left to increase the speed.

From these first principles there is no advantage for either cycling or running; both operate within this constraint. You can't beat physics.

The remaining part of the answer is engineering more than pure physics.

First we need to understand why a bicycle in flat terrain can go faster than a runner can run even though they operate under the same physical constraint. I think that the limiting factor for a runner is the back and forth movement of the legs. Legs have evolved to operate efficiently at normal walking and sustained running speeds when gravity can assist on part of the moving cycle. For high-speed running though, gravity is too slow. The legs must go back and forth faster than they would fall, and a runner must use muscle power to overcome the leg's inertia and increasingly actively accelerate them. While no kinetic energy accumulates (the legs go through the same cycle again and again), muscles create a lot of friction, which can be seen by the sweat we need to dissipate the generated heat. There may also be limits of the force the muscles can generate as well as what the tendons, ligaments and bones can tolerate.

We can get an order of magnitude for that with a back-of-the-envelope calculation. Let's assume that a leg has a mass of 20 kg and the runner is going 36 km/h, or 10 ms/s. That's the speed their feet have relative to the body when they are on the ground. The leg's center of mass, assumed halfway up for simplicity, will then move at 5m/s. This speed has to be reached within a quarter of a cycle (halfway through the forward leg movement as well as the backward movement). With a stride of 1.5m the frequency of a cycle, comprising of two strides, will be $\frac{10m/s}{2*1.5m} = 3 \frac{1}{3} Hz$; the period is $T=0.3s$. The leg must accelerate in a quarter of that (the first half of a half-stride), 0.075s. Its acceleration therefore is $\frac{5m/s}{0.075s} \approx 67m/s^2$. The resulting force on the leg is ${67m/s^2} * 20kg = 1333N$, equivalent to 140kg. (That sounds a bit much — did I make a calculation or estimation error? But perhaps, together with biomechanical advantages like elasticity, cyclic movements etc. it is realistic.)

Bottom line:

Fast running needs a lot of muscle work just to accelerate the legs which loses a lot of energy to heat.

The limiting factor with running are the mechanics of our legs and muscles which limit how fast we can move the legs back and forth.

For bicycling this limitation is bypassed with gears: We can switch to higher gears until wind resistance is so large that we cannot muster more leg force to overcome it, by which point we can only ride faster by faster pedaling so that we "run" into the same problem as the runner.

As an illustration imagine riding in a plain at a low gear that requires pedaling at the same frequency as a runner moves their legs, say one complete cycle/3m. I would assume that it is hard to move the legs faster than maybe 3 Hz for both which seems about right for a runner (Usain Bolt made it 44 km/h) as well as a cyclist pedaling like crazy with almost no resistance.

Now when we run or ride uphill, sustained ascent speed will be so low that the frequency of leg movements will no longer be a limiting factor. It all will come down to the gain in potential energy. And here the bicyclist has an elephant-in-the-room sized disadvantage: The bicycle ;-).

I would assume that with the right gear the rider would have a similar speed to a runner carrying a bicycle — there is no reason why not. Bicycle mechanics and rolling resistance need a bit excess energy but I would assume that the cyclic pedaling has less muscle friction than walking which essentially moves the legs "empty" half the time. Professional pedals, by contrast, have click-in mechanisms for the shoes so that the rider can pull during the up-slope of the pedaling cycle, thus minimizing dead movement. That should make up for mechanical heat losses, but probably not for lifting the bicycle itself.

With these considerations we can now make an estimate at the speed at which the advantage of the bicyclist over the runner should taper off: When the leg movement becomes slow enough that not much muscle action is needed to accelerate them. That should be around the time the necessary acceleration is close to Earth's g, if our assumption holds that normal leg movement has evolved to exploit gravity assists to swing them back and forth when unloaded.

As an estimate, we said that a stride is 1.5m. Over that distance, the leg is accelerated until it hits the floor where it has the relative speed of the runner, before it is lifted and decelerated again while the runner is in the air. Let's assume the actual gravity-assisted acceleration is roughly 1/2 g because the leg does not move vertically but follows some curve we can compute the time t it needs to travel from an upper position to the ground from

$s = 1/2 a t^2$

which we solve for t:

$t = \sqrt{\frac{2s}{a}}$

If we assume $a = 5 m/s^2$ and $s = 0.75 m$ we have $t = \sqrt{\frac{1.5m}{5m/s^2}} = \sqrt{0.3s^2} = 0.54s$. Since this is a quarter of an entire cycle, the period T is about 2s and the frequency about 1/2 Hz1. Each complete cycle, two strides, moves the runner 3m so that we have a speed of 3m/2s or 1.5m/s or 5.4km/h, a very brisk walking speed.

Let's recall that we think the disadvantage of a runner is the leg acceleration beyond the gravity assist, and let's assume that the cyclist always has the perfect gear so that their advantage is not to have to do any acceleration work on their legs.

Then the break-even point for the runner will be when the uphill angle is so steep that at the "natural" gravity assist running speed of 1.5 m/s all work is converted into potential energy through altitude gain (and none is lost to leg acceleration).

The cyclist at this angle would not have any advantage left because they could not go any faster either, because physics (and biology).

We'll calculate the altitude gain/s for a moderate sustained human power output of 100 Watt, and then see to what angle that corresponds at the above 1.5m/s.

Incidentally, I weigh about 1000N. With 100 W (or 100 Nm/s) sustained power output I can therefore climb a rate of 0.1 m/s (you just have to love SI units). That would be 100m in 1000s or 20 minutes or so (seems about right. This should be the climbing rate at which the difference between a cyclist and a runner differs only because of the bicycle weight, making the cyclist about 10% slower.

If we look at the triangle in your diagram and draw the triangle of the slope covered at 1.5m/s (a hypotenuse of 1.5 m) and the maximum sustainable altitude gain per second (the vertical side of 0.1 m) we arrive at an incline of 6.6%.

Even though especially the assumptions about leg mechanics were crude and the leg kinetics were terribly oversimplified the result is not entirely implausible for an average heavy person like me. For a person with lower mass and higher power output the incline could easily be two or three times as steep, for example in the Tour de France.

1 We get a similar approximation by considering the leg a pendulum with the center of mass at the knee, about L=50cm from the hip joint. With Earth's gravity of g, the period T of a pendulum with small amplitude is $T=2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.5}{9.81}} \approx 2\pi*0.22 \approx 1.35s.$ Pendulum calculators which correct for large amplitudes give around 1.5s for angles of 60° off the vertical. In any case it's in the same ballpark as the calculation in the text, crude as both are. A shorter period would indicate a faster break-even point with the cyclist.

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    $\begingroup$ That's only 3.78 degrees, which feels like way too shallow a grade for running to be more efficient. I'm not a competitive cyclist by any means, but if given a choice to race someone up a 3.78 degree incline on foot or on bike, I will choose the bike, even if it is a heavy MTB. $\endgroup$ Jan 13, 2022 at 21:04
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    $\begingroup$ @LawnmowerMan Cycle up 6% for an hour and we talk again. Besides, the underlying assumptions are mere guesses so it may well be 12% or 20%. I was trying to show a line of reasoning. $\endgroup$ Jan 13, 2022 at 21:06
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    $\begingroup$ I'm a moderately trained cyclist, and I can sustain (and have sustained) 200W for an hour. Elite cyclists can sustain >400W for an hour. For the duration of a hill, I've seen strong amateur cyclists sustain >800W for a few minutes. Elite pros can probably put out >1000W for a few minutes (the duration of a short, steep hill.) $\endgroup$
    – Duncan C
    Jan 14, 2022 at 2:28
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    $\begingroup$ @Duncan I was trying to establish a baseline on the back of an envelope. All concrete numbers are guesses or convenient numbers (hey, 0.1 m/s!) , the concrete bio-mechanic assumptions are laughable, but I think I outlined the problem in case anybody wants to flesh it out with proper data. (I'll add that to the post.) $\endgroup$ Jan 14, 2022 at 2:40
  • $\begingroup$ "100m in 20 minutes or so." most people reasonably in shape can walk up faster than that $\endgroup$
    – njzk2
    Jan 15, 2022 at 19:18

You cannot find a formula for this because it depends on dozens of individual properties.

As an example, a skilled bicyclist will benefit from the bicycle more, so will want to use the bicycle at a higher angle.

In the rain, the formula changes. In the dark, the formula changes. So on and so forth.

As an interesting edge case, it is worth pointing out that the gear ratios you can select are limited. You can get trapped on a bicycle using your leg muscles in an inefficient way. When running, you can always take shorter strides. At some angle, I become unable to make forward progress on my bike in its lowest gear setting, and thus it is easy to prove running is faster at that point!

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    $\begingroup$ Why would it change in the dark ? $\endgroup$
    – Hans Wurst
    Jan 13, 2022 at 13:02
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    $\begingroup$ @HansWurst Psychology and neuroscience. You're using lots of sensory input to cycle properly. Some of that sensory input is gone at night. As such, while your legs may still be able to provide the necessary force, you may not be able to innervate your muscles at quite the right time to leverage that force. Just to pick one effect: our sense of balance is much better with our eyes open than it is when our eyes are closed, and balancing a bicycle will be affected by this differently than staying atop our feet. You may be able to train much of that difference out, but its still a difference. $\endgroup$
    – Cort Ammon
    Jan 13, 2022 at 16:13
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    $\begingroup$ For future readers: The answer was written before the OP modified the question to exclude it. $\endgroup$
    – Cort Ammon
    Jan 13, 2022 at 16:26

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