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

Question

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?

Answer 1

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.

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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.

Answer 2

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.

Answer 3

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.

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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.

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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.

Answer 4

If

$F = ma\ \ $ and $\ \ E = \frac{1}{2}mv^2\ \ $ and $\ \ E = F·d$

and you apply the following constraints

• friction losses in wheels and air drag are constant or negligible
• force at the wheel hub is constant

Oberth would play a part. Same force applied, same interval of application, same acceleration, would yield a higher $\Delta E$ because the distance covered during force application would be longer, because $E = F·d$.

How about that force? At the hill's bottom at speed, a cyclist is in high gear. Presuming roughly the same force on the pedal from the rider's foot, this does not translate into same driving force at hub when lower gears are in use. The force in these relations refers to the reaction force of the turning wheel axle pushing forward against the bike frame at the hub, not whatever force is being pedaled up on the pedal crank, which is transformed by the gearing.

Unless the cyclist is on a one-speed, one cannot consider the force applied to the pedals, but only the force applied at the wheel hub.

What about those losses we ignored above?

If Newtonian physics applies to bicycles (it does), Oberth has an effect but it's small compared a cyclist's inability to produce an same force regardless of speed (one reason to rocket-power your bike). Losses from air drag are not ever "constant or negligible", at least in my cycling experience. Travel outside an atmosphere and with zero friction from ground contact makes the Oberth effect significant for forces applied in those conditions. Here on the ground, not so much.

Answer 5

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.