The Lewis Ratchet Redeux: The Lewis Ratchet in Space

 

The Lewis Ratchet Revisited: Two Perspectives, One Principle

https://www.youtube.com/watch?v=OxHIMhTpFYU

When we first imagine the Lewis Ratchet—the idea that you can move a platform by throwing and retrieving a mass inside it—it’s often easiest to picture a raft on water. You throw a medicine ball forward (the “kick”), then gently walk it back (the “reset”), and thanks to friction or drag, the raft creeps ahead a little more each cycle.

That initial, Huck Finn–style image might suggest that friction is absolutely essential. After all, friction is what lets you walk without slipping and prevents the raft from drifting back during the slow reset. But in a more general, frictionless environment—say, a sealed box floating in the vacuum of space—the same fundamental process happens via normal forces, collisions, and hooks or latches that allow internal masses to push off the box’s walls.

This is an example of the idea.  The Vpython Script launches 4 masses and then the Spaceman launches.  The goal was to have the Spaceman collect the 4 masses and jump back but my coding skills aren't all that adept.

https://www.glowscript.org/#/user/fifth.grade.physics/folder/perletutchmallotion/program/LewisRatchet

Two Approaches, One Underlying Physics

1. Friction-Based Ratchet (Raft Example)

   • Where: On water or on land, where friction or fluid drag is significant.
   • Why Friction Matters: It “locks in” each forward impulse. When you do the slow reset, friction opposes that movement, preventing the platform from undoing its gain.
   • Visible Effect: Each fast throw or “kick” shifts the platform forward; each gentle retrieval fails to shift it back as much because friction saps momentum. Over many cycles, the raft moves forward.

2. Collision/Normal-Force Ratchet (Box in Space)

   • Where: In free space, friction is absent, but masses still exert forces on each other through collisions or direct pushes on walls.
   • Why No Friction Is Required: Momentum exchange occurs as the mass collides with a wall or is latched in place. Those internal collisions provide the necessary impulses.
   • Visible Effect: Inside the box, each throw (or jump) shifts the box’s outer reference point. No friction is needed—just the fact that when one part of a closed system moves in one direction, the rest recoils in the opposite direction.

Despite these differing mechanisms (friction vs. collision forces), the core physics is the same: asymmetric internal motions can lead to a net displacement of one part of the system (the “platform,” be it a raft or a sealed box). In both setups, momentum conservation holds. You’re simply shifting momentum around inside the system in such a way that after each cycle, the “platform” portion remains offset from where it began.

Why They’re More Alike Than You’d Think

• Same Momentum Accounting: Whether it’s friction on a raft or a normal force in a box, you’re using internal impulses that push the platform one way and a mass the other way.
• Same Asymmetric Cycle: A strong, quick “kick” and a slow, gentle “reset” ensure that the net effect isn’t fully canceled.
• Same End Result: An incremental shift of the platform in the desired direction, repeated cycle by cycle.

Why They Seem Different

• In Everyday Life, We Notice Friction: It’s tangible, we can feel it, so it’s natural to think of friction as the key enabler.
• In Space, Collisions Dominate: No friction needed. The collision or “push-off” from a wall replaces friction as the momentum transfer mechanism, yet the displacement of the box’s outer reference point is just as real.

The Setup

We have an airtight, rigid box floating in deep space—nothing else around for it to push off against. Inside the box are:

1. The Box (160 kg) + Four Separate Masses (40 kg total).

   • The box weighs 160 kg by itself.
   • Four identical masses of 10 kg each add up to 40 kg.
   • So, “box + four masses” together weigh 200 kg.

2. A Spaceman (100 kg).
   This brave soul can move around inside the box, pick up and throw the 10 kg masses, and even jump from one wall to the other.

Altogether, this makes 300 kg of total mass in the system, and that system is isolated—no outside forces.

Box Dimensions

• Length: 10 meters.
  The distance from one end (the “left wall”) to the other end (the “right wall”) is exactly 10 m.

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What Are We Trying to See?

We want to follow each motion inside the box—every time the spaceman throws a mass or jumps across the interior—and find out:

1. How does the box itself shift in space while these internal motions happen?
2. Where is the box’s “outer reference point” (say, a painted mark on the left wall) after all these actions?
3. Do we violate any physical laws? (Spoiler: No. Momentum conservation still holds firmly.)

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The Steps in the Experiment

We’ll perform a total of six big actions:

1. Throw #1: The spaceman throws one 10 kg mass from the left side to the right side of the box at 10 m/s (relative to the box’s interior).
2. Throw #2: Same as above, but a second 10 kg mass is thrown.
3. Throw #3 & Throw #4: Two more identical throws, each time sending a 10 kg mass to the right wall at 10 m/s.
4. First Jump: Now, with all four masses at the right wall, the spaceman (100 kg) jumps from the left wall to the right wall at 0.01 m/s (again, measured inside the box).
5. Second Jump: The spaceman collects all four masses (making himself effectively 140 kg) and jumps back to the left wall at 0.001 m/s.

We’ll track the box’s recoil each time one of these motions occurs.

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Detailed Breakdown

1. Throw #1

• What Happens:
  The spaceman picks up a 10 kg mass (leaving 30 kg still “attached” to the box plus the box’s own 160 kg = 190 kg, plus the spaceman’s own 100 kg for a moment, etc.) and hurls it at 10 m/s to the right.
• Why the Box Moves:
  By Newton’s Third Law, if the 10 kg mass goes right, the rest of the system (box + spaceman + other masses) gets an equal and opposite momentum to the left.
• Time:
  The mass crosses 10 m at 10 m/s, so it takes 1 second to reach the right wall. During that second, the box moves left a fraction of a meter.
• Result:
  After 1 s, the thrown mass sticks to the right wall. Everything becomes “one piece” again, except the box is now offset a bit from its original position.

2. Throw #2, Throw #3, Throw #4

• Repeated Logic:
  Each throw is exactly the same scenario: a 10 kg mass gets tossed at 10 m/s over a distance of 10 m, taking 1 second each time, imparting another small recoil to the box.
• Cumulative Effect:
  After four throws, all four 10 kg masses are at the right wall, and the box has shifted noticeably to the left (in our coordinate system).

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3. First Jump (Spaceman Alone)

• New Setup:
  Now, the four masses are attached to the right wall. This means the box “and those masses” weigh 200 kg together, while the spaceman (100 kg) is still on the left side.
• The Jump:
  The spaceman pushes off the left wall at 0.01 m/s, heading to the right.
• Time:
  He takes $\tfrac{10\,\text{m}}{0.01\,\text{m/s}} = 1000$ seconds to drift across.
• Box Recoil:
  During these 1000 seconds, the 200 kg box will move the opposite way (left) at a small speed (a few thousandths of a meter per second). Multiply that small speed by 1000 seconds, and you get a few meters of displacement.
• End State:
  The spaceman meets the right wall, “sticks” there (again, effectively no momentum left in the system), and the box ends up even further to the left in the inertial frame than before.

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4. Second Jump (Spaceman + 40 kg)

• Now the Spaceman Picks Up the Masses:
  That means he (100 kg) plus the four 10 kg masses (40 kg) become a 140 kg lump. The remaining box alone is 160 kg.
• He Jumps Left at 0.001 m/s:
  This is an even slower internal speed, but the crossing is still 10 m, so it’ll take 10,000 seconds.
• Box Recoil:
  The 160 kg box moves to the opposite side at a tiny speed for 10,000 seconds. Over that long time, it can shift another several meters left.
• Final State:
  By the time the spaceman + masses dock with the left wall, the entire system is back at rest. The outer reference point of the box has ended up significantly displaced from the original zero mark. (In one typical numeric example, it might be around –9 m.)

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What We Learn

1. Momentum Conservation Still Reigns Supreme.
   Even though the box’s marker shifts by many meters, the overall center-of-mass of the 300 kg system never budges from its original location.
2. No External Forces Required.
   These internal movements alone rearrange who’s moving and when. Collisions and normal forces between the box walls, the spaceman, and the masses suffice to transfer momentum internally.
3. Yes, the Box Does End Up Far from Where It Started.
   That’s entirely consistent with everything we know from Newton’s laws—because the “box” is just one piece of the puzzle, not the entire center-of-mass.

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In this experiment, the spaceman and masses provide a series of internal “kicks” and “resets.” Each throw or jump imparts a recoil to the box, and the net effect, after all six actions, is a final offset of the box’s reference point by several meters.

It’s a vivid demonstration that internal motions don’t change the total momentum of the system (the center-of-mass stays put), but they can produce a large displacement of one part of the system—namely, the box we’re looking at—so long as we carefully manage each collision and reattachment inside.

Now we can look at the math and the derived equations that will make working with the Lewis Ratchet easier:

1. Single-Event Derivation

Consider a closed system of total mass $M_{\mathrm{tot}}$ in free space. At some step, a sub‐mass $m$ (which might be one of the small thrown masses, or the spaceman, or the spaceman + other masses) moves from one wall of the box to the other at speed $v_{\mathrm{rel}}$ relative to the box. The “rest of the system” has mass $M$, such that

$$m + M = M_{\mathrm{tot}}\quad (\text{for that instant}).$$

(If the mass $m$ is “throwing itself” out of the rest, then $M$ is everything still attached to the box.)

1.1. Momentum Conservation

Right after the push/throw, in the inertial frame:

• The “moving mass” $m$ has velocity $\bigl(v_{\mathrm{box}} + v_{\mathrm{rel}}\bigr)$.
• The rest $M$ (the box + everything attached) has velocity $v_{\mathrm{box}}$.

Since the total momentum was zero before the event (assuming the whole system was momentarily at rest), we have

$$M\,v_{\mathrm{box}} \;+\; m\,\bigl(v_{\mathrm{box}} + v_{\mathrm{rel}}\bigr) = 0.$$

Rearrange to solve for $v_{\mathrm{box}}$:

$$M\,v_{\mathrm{box}} \;+\; m\,v_{\mathrm{box}} \;+\; m\,v_{\mathrm{rel}} = 0 \;\;\Longrightarrow\;\; (M + m)\,v_{\mathrm{box}} = -\,m\,v_{\mathrm{rel}} \;\;\Longrightarrow\;\; \boxed{ v_{\mathrm{box}} = -\,\frac{m\,v_{\mathrm{rel}}}{M + m} }.$$

(Sign convention: if $m$ moves to the right at $v_{\mathrm{rel}}$, the box recoils to the left.)

1.2. Transit Time and Box Displacement

Inside the box, the sub‐mass $m$ crosses distance $L$ at speed $v_{\mathrm{rel}}$. The time of travel is

$$t =\frac{L}{v_{\mathrm{rel}}}.$$

During that time, the box moves at constant velocity $v_{\mathrm{box}}$. Hence the box’s displacement $\Delta x$ is

$$\Delta x = v_{\mathrm{box}} \times t = \Bigl(-\,\frac{m\,v_{\mathrm{rel}}}{M + m}\Bigr)\; \Bigl(\frac{L}{v_{\mathrm{rel}}}\Bigr).$$

Notice the $v_{\mathrm{rel}}$ cancels:

$$\Delta x = -\, \frac{m\,L}{M + m}.$$

This is the net shift of the box’s outer reference point in the inertial frame during that single crossing. Once $m$ sticks to the opposite wall (an inelastic collision that brings it back to the same velocity as the box), the system is again at rest overall—but offset by $\Delta x$.

Lewis Ratchet Equation (Single Event)

$$\boxed{ \Delta x \;=\; -\,\frac{m\,L}{M + m} }.$$

• The negative sign indicates the box recoils opposite to the mass’s internal travel direction.
• No external force is involved, so the center-of-mass remains fixed in space; we’re only tracking the box’s marker (one piece of the system).

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2. Multiple Events: Summation for Throws and Jumps

In the Lewis Ratchet thought experiment, you often have a sequence of internal motions:

1. Throw #1: A 10 kg mass moves from left to right at 10 m/s (relative speed).
2. Throw #2, #3, #4: Additional masses thrown similarly.
3. Jump #1: The spaceman (perhaps 100 kg alone) jumps from one wall to the other.
4. Jump #2: The spaceman + any collected masses (could be 140 kg) jumps back.

Each step can be treated with the same formula, but note that $m$ and $M$ change as masses detach or reattach. The basic pattern remains:

• If a sub-mass $m_i$ moves across distance $L$ at speed $v_{\mathrm{rel},i}$ inside the box, the box’s displacement in that event is

$$\Delta x_i = -\, \frac{m_i\,L}{M_i + m_i},$$

where $M_i$ is whatever remains attached to the box at that moment.

• After each event, the mass $m_i$ may reattach, so the next event uses a new mass split $(m_{i+1}, M_{i+1})$.

Hence the total displacement of the box after $N$ internal events is just

$$\Delta X = \sum_{i=1}^{N} \Delta x_i = \sum_{i=1}^{N} \Bigl( -\,\frac{m_i\,L}{M_i + m_i} \Bigr).$$

(Time durations, i.e. crossing times $t_i = L / v_{\mathrm{rel},i}$, set how long each recoil velocity lasts, but ironically you don’t even need $v_{\mathrm{rel},i}$ in the final expression, since it cancels out.)

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3. Interpretation and Significance

1. No External Forces:
   All momentum exchanges are strictly internal, so the center‐of‐mass does not move in the inertial frame. The box’s marker, however, can end up far from its start—because “the box” is not the entire system, just one component.

2. General Applicability:

   • Throws: If a small mass $m$ is “fired” to the right at relative speed $v_{\mathrm{rel}}$, the box of mass $M$ recoils left.
   • Jumps: If the “moving sub‐mass” is the spaceman or spaceman+other masses, the formula is the same. The sign changes depending on which way he/they push off, but the magnitude is always $\tfrac{m\,L}{M + m}$.

3. Why It Looks Surprising:
   Intuition might say, “You can’t move the box by shuffling stuff inside.” But the equation shows the box’s outer reference point indeed shifts each time. Meanwhile, the overall center-of-mass stays fixed, preserving momentum conservation.

4. No Violation of Physics:
   This formula is precisely what momentum conservation predicts once you account for the mass that’s moving relative to the rest of the system. No exotic forces or external pushes are needed—just internal impulses and the finite size of the box.

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Final Word

Thus, the Lewis Ratchet “equation” for a single crossing is

$$\Delta x = -\,\frac{m\,L}{M + m},$$

and for multiple sequential events, you simply sum up each $\Delta x_i$. This straightforward relation is the mathematical heart of why the box’s reference point can end up significantly displaced—yet with zero net external momentum. It’s a powerful illustration of how internal motions in a closed system still allow one part of that system (the box’s marker) to move, so long as other components compensate in just the right way to keep the total center‐of‐mass unchanged.

For those of you concerned with violations of the Center of Mass principle:

The Lewis Ratchet: A Focus on Center of Mass

Introduction

Ever heard someone say, “You can’t move a box in space by shuffling things around inside it”? Well, the Lewis Ratchet demonstrates that you can shift the box’s outer shell by using internal throws and jumps—yet this still obeys all the usual physics laws. Let’s break down exactly what happens and why it makes perfect sense once you understand momentum conservation and the center‐of‐mass.

The Setup

1. Box + Masses = 200 kg
   • The box itself weighs 160 kg.
   • There are four separate 10 kg masses (total 40 kg).
2. Spaceman = 100 kg
3. Total system = 300 kg
4. Box length = 10 m
5. Throw speed = 10 m/s (relative to the box)
6. Two Jumps for the spaceman:
   • First at 0.01 m/s (spaceman alone)
   • Second at 0.001 m/s (spaceman carrying all four masses)

Initially, everything is at rest in an inertial frame, and the outer reference point of the box is at position $x=0$. We analyze each throw or jump one step at a time.

Four Throws

• Action: The spaceman throws a 10 kg mass from one side of the box to the other, at 10 m/s relative to the box.
• Consequence: The box (and whatever remains attached to it) recoils a bit in the opposite direction.

In each of the four throws:

1. A 10 kg mass is launched.
2. The mass travels 10 m in 1 second (because 10 m ÷ 10 m/s = 1 s).
3. Meanwhile, the recoil moves the box’s outer reference point a fraction of a meter in that one second.
4. Once the mass sticks to the far wall, the system again becomes a single unit—but now positioned slightly offset from where it started.

By the end of these four throws, the box’s outer reference point has shifted by over a meter. Throughout these internal motions, total momentum stays zero for the entire 300 kg system—but the box’s shell has moved.

First Jump (Spaceman Alone)

Next, the spaceman (100 kg) jumps across the 10 m interior at 0.01 m/s:

• Time: 10 m ÷ 0.01 m/s = 1000 s.
• Recoil: While he’s jumping, the 200 kg box‐plus‐masses moves in the opposite direction at a small speed.
• Net Result: Because 1000 s is substantial, that small recoil speed shifts the box’s outer reference point a few more meters.

When the spaceman lands and reattaches, the system again becomes a single mass, momentarily at rest—but now the box is displaced even further.

Second Jump (Spaceman + 40 kg)

Finally, the spaceman picks up the four 10 kg masses, becoming a combined 140 kg “lump.” He jumps back at 0.001 m/s:

• Time: 10 m ÷ 0.001 m/s = 10,000 s.
• Recoil: The 160 kg box creeps in the opposite direction at an even smaller speed.
• But: 10,000 s is a lot of time, so even that tiny recoil speed translates to several more meters of displacement.

In the end, the entire 300 kg system is again assembled into one piece and at rest, only now the box’s reference point can be many meters away from where it started.

The Core Reasoning: Center‐of‐Mass vs. Box Marker

• Center‐of‐Mass Stays Put: Through every throw and jump, the system’s overall momentum is zero. No external forces act, so the center‐of‐mass doesn’t move from its original position in the inertial frame.
• Box Marker Moves: The “shell” or “marker” on the box is not the center‐of‐mass. As internal components rearrange and swap momentum, the box shell can wind up several meters from its original spot. That’s the entire point of the Lewis Ratchet: internal actions can create relative motions within the system, leaving a visible shift in the box’s outer wall even though no external push has been added.

Final Thoughts

The Lewis Ratchet shows, step by step, how a spaceship or box in free space can end up displaced through entirely internal maneuvers—but there’s no mystery or violation of physics. The math is straightforward, and momentum conservation is never broken:

1. Each throw or jump redistributes momentum inside.
2. The system ends at rest again, offset in position, because the center‐of‐mass remains fixed and the box’s wall is simply not the same as the center‐of‐mass location.

This is all absolutely consistent with Newton’s laws. Observing the box’s outer reference point alone can fool you into thinking it “moved for free,” but if you track everything together, you’ll see that the center‐of‐mass never left its initial coordinates.

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In essence, the Lewis Ratchet is a fascinating illustration of how internal throws and jumps can produce large final displacements of a box’s exterior, all without violating the fundamentals of classical mechanics. It’s a cool puzzle—one that reaffirms just how elegantly Newtonian physics handles “internal vs. external” forces and how the center‐of‐mass is the true anchor of momentum conservation.