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When faced with a series of ups and downs while riding a bicycle, I try to pedal like mad near the bottoms to gain as much speed as possible. I do this due to some vague, ill-formed notion that either I'm taking advantage of something like the Oberth effect, or that I'm reducing gravity drag.

If I were doing the former I'd pedal hardest for a period centered around the lowest point between hills, but if I were doing the latter I'd pedal hardest when the uphill section had maximum slope.

Perhaps neither is correct, but one is the best analogy.

Question: Is biking up and down a series of hills a good real-world analogy for understanding either the Oberth effect and/or gravity drag? Which is a better analogous match from a mathematical perspective?

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    $\begingroup$ Isn't this question more suited on Physics SE or even Bicycle SE? $\endgroup$ – asdfex Aug 19 '20 at 11:20
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    $\begingroup$ @asdfex I think this is on-topic and suitable for either site, in that case OP simply chooses the community they'd prefer to receive answers from and where they think it will fit in with other posts. I'm trying to find if the bicycle analogy can be used to help explain Oberth or gravity drag, and I feel this will be of more use here in Space SE where both are oft-discussed topics. I think if nobody can muster an answer here that's supported by math, I'll post one myself just to make sure the results are conclusive. $\endgroup$ – uhoh Aug 19 '20 at 11:23
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    $\begingroup$ This would work if air drag wasn't a viscous friction (proportional to velocity). It would work great in conditions where friction is low, but on bike increasing peak velocity will increase the losses the most. $\endgroup$ – SF. Dec 21 '20 at 10:43
  • $\begingroup$ @SF. I think so too, that's why the word "frictionless" appears in the bounty message. Since I don't bike very fast but use a nice bike, most of the time "frictionless" is a good approximation for me as well. $\endgroup$ – uhoh Dec 21 '20 at 11:37
  • $\begingroup$ @uhoh I think a more intuitive example would be a car moving on flat surface, with friction. You start from a stop, and you are allowed to floor the accelerator twice, for 3 seconds every time, choosing the moments yourself. Floor it once, come to a stop, repeat, come to a stop. Check the distance. Now repeat, but just hold the accelerator for 6s from the beginning. The time it takes until you stop is the same (or even shorter) as the two segments in the first case, but you'll cover much more distance, because during the first half of the time you're moving at a significantly higher speed. $\endgroup$ – SF. Jan 8 at 5:23
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No, Oberth can't bike.

Why you want to "pedal like mad" in the brief downhill before an uphill, intuitively, is because your physiology is constrained by peak power, so you don't want to waste that by slacking off downhill.

The Oberth effect is motivated not by a power constraint, but by a fuel constraint. If you wanted to bicycle in a way that simulated the Oberth effect, you'd fast for a few days beforehand, eat one plate of pasta, and then mete that out over a century (cyclist jargon for a hundred miles) without bonking (cyclist for I'm a depleted ni-cad). Which doesn't work because a cyclist's resting metabolism as a fraction of peak output (100 W vs 1000 W) is way more than a spacecraft's (300 W vs 30000 W?). As mammals go, even the sloth's legendarily small fraction is enormous compared to something that can drift along on solar panels while fuel just sits in its tanks.

I don't have hard numbers for 300 vs 30000, but that's the route to a mathematical answer. Saturn V's third stage J-2 put out 7800 hp (page 4 of this summary), 5850 kW; at that point, the stack's idle power usage must have been closer to one thousandth of that than a cyclist's one tenth.


To simulate different bicycling strategies, one could write a short program (these days, probably in Python) to measure figures of merit such as m/s or m/J. Model a stretch of hilly road as a sum of sinusoids. Choose the cyclist's mass, sustained power, and sprinting power. Estimate the cyclist's drag w.r.t. speed (always zero, like a spacecraft?). Simulate the journey at a time step of a second or so. Vary when to sprint: when speed exceeds a threshold, or during the ten seconds before hill angle exceeds a threshold, or even when speed drops below a threshold (climbing out of the saddle) -- which would be like the opposite of the Oberth effect. For fairness, somehow arrange that all strategies sprint for roughly the same total duration.

If you care only about m/J, like a fuel-constrained spacecraft instead of a racing cyclist, then coasting at the 100 W metabolic baseline should also be allowed. Then the optimal strategy is simply to coast whenever speed exceeds some fairly low threshold, and probably never to sprint.

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    $\begingroup$ Thank you for your thoughtful answer and delicous analogy! While this prose-based answer is interesting, it "goes to motive and opportunity" rather than to the underlying mathematics. Regardless of why one would or would not want to use the Oberth effect,* mathematics still shows* that a given impulse will raise the energy of an orbit more if it occurs at periapsis than at apoapsis. If it's also possible to address what's written in the bounty message directly, that would be great! Nice hat by the way, it suits you :-) $\endgroup$ – uhoh Dec 20 '20 at 5:22
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    $\begingroup$ I regret that my math isn't up to that. But my hunch is that, as an analogy to spacecraft propulsion, cycling is like an ion thruster with a huge tank of xenon and such a low thrust that you might as well keep the pedal (ahem) floored all the time. $\endgroup$ – Camille Goudeseune Dec 20 '20 at 5:31
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The reason your biking scheme feels easier is, because the power you put into the pedals is applied for a longer time and therefore lower. There is no relation to Oberth effect because the total energy spent is constant.

Compare the two cases:

  • Pedal only going up-hill - you have to apply power during the time going uphill.
  • Pedal going up-hill and in the plane - you apply power during a longer period and lose speed while going uphill. In the straight section you gain kinetic energy you then can spend going uphill.

The total amount of energy is about the same. There are 3 components that contribute to the total energy spent:

  • Potential energy. This is always the same as the height difference is not changed
  • Losses due to friction. This scales with the velocity squared, so in your approach losses are higher because your speed is higher.
  • Losses due to down-hill forces. In order not to roll backwards, you have to apply some force to the pedals which costs your body some energy to produce. This energy is only needed due to the particular way your body produces force in its muscles. It would be sufficient to just put a weight on the pedals to counteract this force. The energy your body has to spend is changes with the time you stay on the slope (inversely proportional to speed), so is lower using your approach. This is roughly equivalent to gravity drag.

In sum, the total energy spend is likely higher using your approach due to larger air resistance. Nevertheless, it might feel easier because the maximum power you need to apply is lower.

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  • $\begingroup$ Just because the total kinetic+potential energy the cyclist receives from pedaling is the same in two cases (up to losses to friction and air resistance), doesn't mean that the total chemical energy spent by the muscles is the same. Our muscles can spend energy even when performing no work at all, like pushing a wall or holding a weight in an outstretched hand. And it's actually similar to Oberth's effect: for example, a spacecraft in an elliptic orbit, in order to get to an escape trajectory, needs the same amount of energy in any point of the orbit; (cont) $\endgroup$ – Litho Aug 20 '20 at 7:41
  • $\begingroup$ ...however, the amount of fuel's stored chemical energy the spacecraft's engines would have to spend depends on the point of orbit where the burn is performed. $\endgroup$ – Litho Aug 20 '20 at 7:44
  • $\begingroup$ @Litho "Our muscles can spend energy even when performing no work at all" - That's exactly what I described and the reason why I split the total energy in three parts. Was my explanation not clear enough? But this fact is the equivalent to gravity drag (fuel is spent to keep the rocket in place), but has nothing to do with Oberth. $\endgroup$ – asdfex Aug 20 '20 at 8:02
  • $\begingroup$ Sorry, I missed that part of your explanation. However, it seems to me that existence of these losses implies that there's likely an analog of Oberth effect as well. What I mean is this: we see that the amount of energy the cyclist's muscles spend depends not only on the amount of work they perform, but also on the amount of momentum they transfer. If we assume that the dependence on momentum is more or less independent of cyclists's speed (though I admit that this is far from obvious), then... (cont.) $\endgroup$ – Litho Aug 20 '20 at 8:41
  • $\begingroup$ ... then the corresponding part of spent energy is smaller when the cyclist receives the same amount of kinetic+potential energy at higher speed than at lower speed. $\endgroup$ – Litho Aug 20 '20 at 8:43
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While the actual efficiency of this strategy is questionable and involves biomechanics, friction, and many other complicating factors, the most reduced model can indeed be compared to the Oberth effect.


What you are "feeling" is the force you are applying. "Hard" is when it takes you a lot of force to move the pedals, "easy" is when it doesn't. Your body can be approximated to being capable of applying a certain force, and when it's "easy" to pedal, you can just pedal harder in order to reach your standard level of force.

In this view, you're no different from a rocket engine, which also applies a certain force to the spacecraft.

The Oberth effect, in its core, is about applying a force in the same direction as you are travelling, at the highest possible velocity to maximise energy the energy gain.

In the same way, the bottom of the hill is where your velocity is the highest, and an applied force would add the most energy.


This isn't necessarily a useful observation, since the forces slowing you down are also bleeding away a larger amount of energy at higher speeds.

Is biking up and down a series of hills a good real-world analogy for understanding either the Oberth effect and/or gravity drag?

I don't think so. If you understand the Oberth effect, you could point out the similarity. But you are in effect explaining a simple scenario with a more complicated one. "gravity", as explained by "biomechanics", "friction" and "gravity".

Analogies are useful when they can replace a difficult idea with an easier idea.

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No. The Oberth effect is purely a space thing.
Or, more precisely, it is an effect of propelling yourself by hurling reaction mass backwards.

I think, this is best explained using a little thought experiment. Consider a person on a skateboard. The person weights 70kg and holds a weight of 1kg in their hand. The person hurls the weight backwards at a speed of $\Delta v_e = -7\frac{m}{s}$ to increase their own speed by $\Delta v_r = 0.1\frac{m}{s}$. The person does a work of

$$\Delta E_{kin} = \frac{1}{2}(1kg\cdot v_e^2 + 70kg\cdot v_r^2) = 24.85J$$

Now, lets calculate the energy $E_{e0}$ of the weight and the energy $E_{r0}$ of the skater before hurling, as well as the energy $E_e$ of the weight and the energy $E_r$ of the skater after hurling. Finally, calculate the $\Delta E = E_e + E_r - E_{e0} - E_{r0}$ of the entire system and the $\Delta E_r = E_r - E_{r0}$ of the skater. I do this for three different cases:

  1. The skater is at rest before hurling.

    $E_{e0} = 0J$
    $E_{r0} = 0J$
    $E_e = 24.5J$
    $E_r = 0.35J$
    $\Delta E = 24.85J$
    $\Delta E_r = 0.35J$

  2. The skater moves at $7\frac{m}{s}$ before hurling.

    $E_{e0} = \frac{1}{2}1kg\cdot (7\frac{m}{s})^2 = 24.5J$
    $E_{r0} = \frac{1}{2}70kg\cdot (7\frac{m}{s})^2 = 1715J$
    $E_e = 0J$
    $E_r = \frac{1}{2}70kg\cdot (7.1\frac{m}{s})^2 = 1764.35J$
    $\Delta E = 24.85J$
    $\Delta E_r = 49.35J$

  3. The skater moves at $20\frac{m}{s}$ before hurling.

    $E_{e0} = \frac{1}{2}1kg\cdot (20\frac{m}{s})^2 = 200J$
    $E_{r0} = \frac{1}{2}70kg\cdot (20\frac{m}{s})^2 = 14000J$
    $E_e = \frac{1}{2}1kg\cdot (13\frac{m}{s})^2 = 84.5J$
    $E_r = \frac{1}{2}70kg\cdot (20.1\frac{m}{s})^2 = 14140.35J$
    $\Delta E = 24.85J$
    $\Delta E_r = 140.35J$

You see, even though the work done by the skater $\Delta E$ is always the same, the gain in kinetic energy by the skater $\Delta E_r$ depends heavily on its speed. The difference comes from the amount of kinetic energy that is removed from the weight as it is hurled backwards. This energy ends up as kinetic energy of the skater.

The Oberth effect is that the faster a rocket is going, the more kinetic energy is redistributed between the rocket and the fuel, adding to the change of the rocket's kinetic energy.

Bikes obviously work very differently: Their reaction mass is effectively infinite and always unmoving in the reference frame of the earth (as the reaction mass is earth itself). As such, a biker cannot remove energy from their reaction mass to take advantage of the Oberth effect. Instead, the $\Delta v$ that a biker gains from expending a fixed amount of energy goes down as their speed increases.

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  • $\begingroup$ Disagree! Oberth effect applies to the use of any delta-v or impulse and has nothing to do with whether it's due to reaction mass or not. If you could use a solar sail for five minutes per orbit you'd get the biggest bang at the highest velocity point of your orbit. That's why impulse from pedaling is actually a good rather than a bad analogy. Reaction mass makes the calculation harder unless the mass change is so small that you ignore it in the calculation. $\endgroup$ – uhoh Dec 21 '20 at 11:32
  • $\begingroup$ @uhoh Well, at first you got me thinking "oh, there seems to be a hole in my reasoning". I had indeed not considered light sails. However, even with a light sail, you are hurling reaction mass backwards. In the form of photons. And you are doing that at a specific speed, the speed of light. And you are also taking your kinetic energy from the photons, for they will look red-shifted to the observer at rest (or blue shifted while braking). The longer I think about this, the more I'm convinced that my answer is spot on even in this corner case. But thanks for challenging me to think deeper :-) $\endgroup$ – cmaster - reinstate monica Dec 21 '20 at 18:50
  • $\begingroup$ Though they carry momentum, photons do not have mass, so reflecting light is not really "hurling reaction mass". Impulse is a concept in mechanics applicable anywhere, it's not a "space thing". $\endgroup$ – uhoh Dec 21 '20 at 21:48
  • $\begingroup$ @uhoh Well, the photons do carry a mass which is equal to their energy. They just don't have a rest mass (or rather they cannot move slower than light speed, which is the same in relativity). That's why photons redshift when they loose energy/mass. And it is the interaction with this mass that transfers momentum to the solar sail. Also, the overall energy/mass of the photons, the solar sail, and the space probe must remain constant; so the kinetic energy that the probe gains/looses must come from a redistribution of energy between photons and probe. $\endgroup$ – cmaster - reinstate monica Dec 21 '20 at 22:02
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    $\begingroup$ No, it is not a "space thing" and no hurling is necessary. Oberth derivation requires only a force integrated over a path, or in its simplification, an impulse. There is no need at all to specify the mechanism behind the force or impulse, it's just that historically internally stored reaction masses have been used, so they provided a familiar framework. It's a legacy explanation, but absolutely not a necessary part of the derivation or experimental realization of the Oberth effect. $\endgroup$ – uhoh Dec 23 '20 at 1:04

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