The Lewis Ratchet: Forward Asymmetric Momentum Transfer

 

The Lewis Ratchet: Forward Asymmetric Momentum Transfer

I figured out how to make a model of this: https://www.youtube.com/watch?v=A5x0Yd7kjSc

Imagine you’re floating on a small raft in a calm lake—a bit like Huck Finn’s raft, but with a twist. Instead of a motor or engine, your raft relies on an ingenious idea: by moving an internal mass back and forth in an asymmetric way, you can slowly “scooch” the raft forward. This is the essence of the Lewis Ratchet.

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The Basic Idea

In an ideal, frictionless world, any internal motion—like shaking a heavy bar or throwing a medicine ball—cancels out. When you push one way, you push the other equally. But if you can arrange for the forward push to be strong and quick, and the return motion to be gentle and slow, the momentum transferred during the fast “kick” isn’t completely undone by the slow “reset.” In a real-world setting, friction (or water drag) then “locks in” that net gain, causing the raft to move forward incrementally with each cycle.

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A Simple Thought Experiment

The Throw:

Imagine you have a heavy medicine ball on your raft. You throw it forcefully toward a wall on the raft. The rapid acceleration of the ball generates a strong backward impulse. By Newton’s third law, the raft gets an equal and opposite forward impulse.

The Reset:

Now, you need to get the ball back. But instead of throwing it back with the same force, you walk slowly to retrieve it. Your gentle, measured motion imparts a much smaller impulse on the raft.

The Net Effect:

Because the forward kick (from throwing the ball) is stronger than the gentle reset (from walking to pick it up), the raft ends up shifting slightly forward. Over many cycles, these small shifts add up to noticeable movement.

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The Mechanics of the Lewis Ratchet

The key to the Lewis Ratchet is how momentum is redistributed within the system. Consider a simplified model:

• Let m be the mass you move (for example, the medicine ball or a heavy bar).
• Let M be the mass of the raft (plus you).
• The total mass is M_total = M + m.

When you move the mass by a distance D relative to the raft, the raft itself shifts by an amount proportional to the ratio m/M_total. In a perfectly symmetric cycle (fast forward, slow back with equal displacements), the net movement would be zero. But if you design the cycle so that the forward displacement (D_forward) is larger than the return displacement (D_return), the net displacement per cycle is

$$\mathrm{\Delta }{x}_{\text{net}}=\frac{m}{M+m}(D_{\text{forward}}-D_{\text{return}})\mathrm{.}$$

This equation tells you that even if the internal mass moves back and forth, as long as there’s an asymmetry in how far (or how fast) it moves in each phase, you get a net shift.

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Why It Works on the Ground or in Water

In a frictionless environment like deep space, any internal motion cancels out perfectly—the forward and reverse impulses are equal and opposite, and the system’s overall center of mass remains unchanged. But on a surface with friction or water drag, energy is dissipated during the reset phase. That dissipation “locks in” the forward gain from the kick phase. In other words, the friction prevents the raft from “bouncing back” completely when you reset the mass. You’re not violating momentum conservation—you're simply taking advantage of real-world forces that prevent perfect cancellation.

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Distinguishing Between the Lewis Ratchet and CLPP

It’s important to understand how the Lewis Ratchet differs from Closed Loop Pulse Propulsion (CLPP):

• The Lewis Ratchet is Linear:
  It uses straight-line, asymmetric motion of an internal mass. The mechanism works by rapidly moving the mass in one direction (a strong “kick”) and then slowly returning it (a weak “reset”), producing a net forward shift. This linear motion only works effectively when there’s friction or another external dissipative force (like water drag) to “lock in” the gain.

• CLPP Uses Angular Momentum:
  In contrast, CLPP relies on converting linear momentum into angular momentum and vice versa. By carefully managing the redirection of momentum—often using rotating elements or pulses that cycle through angular states—CLPP can produce continuous thrust even in near-frictionless space. CLPP is designed to overcome the inherent cancellation of internal motions in an isolated system, making it a candidate for propellant-less space propulsion.

In short, while the Lewis Ratchet demonstrates that internal, asymmetric motions can generate net movement (ideal for ground or water where friction is available), CLPP leverages angular momentum and precise momentum redirection to create sustained thrust in space. Both concepts hinge on the principle of asymmetric impulse transfer, but they operate in very different regimes.

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The Takeaway

The Lewis Ratchet shows that if you can move an internal mass in an asymmetric way—pushing hard and resetting gently—the net effect is movement. On a frictional surface, this “ratchet” effect becomes cumulative, causing your raft (or any platform) to slowly shift its position. While it’s not a method for space propulsion (because, without friction, internal impulses cancel), it lays the conceptual groundwork for more advanced ideas like CLPP, where angular momentum plays the central role.

Each cycle adds a little bit to the total momentum. Over many cycles, even a small net impulse can lead to measurable movement—a “scooch” or “swish” across the ground or water.

Closing Thought:

The Lewis Ratchet shatters a long-held assumption—that you must push against water, air, or any external medium to change your position in space. It proves that you can shift your x, y, and z coordinates through nothing more than a cleverly unbalanced, internal motion. It’s almost ridiculously simple: a fast, forceful “kick” paired with a gentle, dissipative reset can, over time, ratchet your entire system forward. This isn’t magic—it’s a fundamental rethinking of movement. In a world where we once believed such repositioning to be impossible without an external reaction, the Lewis Ratchet reminds us that sometimes the simplest ideas have the power to redefine the limits of what we thought was achievable.

Addendum:

Below is a rigorous, step‐by‐step mathematical addendum that “proves” the basic principle behind the Lewis Ratchet. This derivation shows that if you move an internal mass along a fixed track relative to a platform by an asymmetric distance in the forward and reverse phases, the platform’s center-of-mass will shift by a net amount. (Note that, strictly speaking, in a frictionless, perfectly isolated system the internal motions cancel; here we assume that friction or damping “locks in” the displacement so the net shift is maintained.)

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The System

Consider an isolated system composed of two parts:

• A platform of mass $M$ (which may include you and the raft), and
• An internal mass $m$ (for example, a medicine ball or heavy bar) that is free to move relative to the platform along a linear guide.

The total mass is

$$M_{\text{total}}=M+m\mathrm{.}$$

In the absence of external forces, the center of mass (COM) of the system is given by

$$X_{\text{COM}}=\frac{M{X}_M+m{X}_m}{M+m},$$

and $X_{\text{COM}}$ remains constant.

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Displacement of the Internal Mass

Assume initially the platform and the internal mass are at positions $X_M^0$ and $X_m^0$, respectively, with

$$X_{\text{COM}}=\frac{M{X}_M^0+m{X}_m^0}{M+m}\mathrm{.}$$

Now suppose you move the internal mass relative to the platform by a displacement $D$ in the forward (or “kick”) phase. Because the internal mass moves relative to the platform, its new position relative to an inertial (water- or ground-fixed) frame is:

$$X_m^{\mathrm{\prime }}=X_M^{\mathrm{\prime }}+D,$$

where $X_M^{\mathrm{\prime }}$ is the new position of the platform.

To keep the COM unchanged, the new positions must satisfy

$$\frac{M{X}_M^{\mathrm{\prime }}+m(X_M^{\mathrm{\prime }}+D)}{M+m}=\frac{M{X}_M^0+m{X}_m^0}{M+m}\mathrm{.}$$

If we assume that before the motion the internal mass was at the same location as the platform (or that we define the relative displacement from the starting point as zero), we can set $X_m^0=X_M^0$ so that initially the COM is simply $X_M^0$.

After the displacement, we have:

$$\frac{M{X}_M^{\mathrm{\prime }}+m(X_M^{\mathrm{\prime }}+D)}{M+m}=X_M^0\mathrm{.}$$

Multiply both sides by $M+m$:

$$M{X}_M^{\mathrm{\prime }}+m{X}_M^{\mathrm{\prime }}+mD=(M+m)X_M^0\mathrm{.}$$

Factor $X_M^{\mathrm{\prime }}$:

$$(M+m)X_M^{\mathrm{\prime }}+mD=(M+m)X_M^0\mathrm{.}$$

Subtract $(M+m)X_M^{\mathrm{\prime }}$ from both sides:

$$mD=(M+m)(X_M^0-X_M^{\mathrm{\prime }})\mathrm{.}$$

Solve for the displacement of the platform:

$$X_M^0-X_M^{\mathrm{\prime }}=\frac{m}{M+m}D\mathrm{.}$$

Thus, the platform moves in the opposite direction by

$$\mathrm{\Delta }{x}_{\text{forward}}=\frac{m}{M+m}D\mathrm{.}$$

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Asymmetric Cycle: Forward Kick and Slow Reset

Now, suppose the cycle is performed in two phases:

1. Forward (Kick) Phase:
   The internal mass is moved quickly by a distance $D_{\text{forward}}$ relative to the platform. By the derivation above, the platform shifts by

   $$\mathrm{\Delta }{x}_{\text{forward}}=\frac{m}{M+m}{D}_{\text{forward}}\mathrm{.}$$

   (This impulse is large because the motion is fast and forceful.)

2. Return (Reset) Phase:
   The internal mass is then returned slowly by a distance $D_{\text{return}}$ (measured relative to the platform) so that the reverse impulse is much smaller. During this slow phase, the platform will move in the opposite direction by

   $$\mathrm{\Delta }{x}_{\text{return}}=\frac{m}{M+m}{D}_{\text{return}}\mathrm{.}$$

Because the forward phase is designed to be much more forceful (or covers a larger effective displacement) than the return phase, we have

$$D_{\text{forward}}>D_{\text{return}}\mathrm{.}$$

The net displacement of the platform per cycle is therefore:

$$\mathrm{\Delta }{x}_{\text{net}}=\mathrm{\Delta }{x}_{\text{forward}}-\mathrm{\Delta }{x}_{\text{return}}=\frac{m}{M+m}(D_{\text{forward}}-D_{\text{return}})\mathrm{.}$$

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Final Result

This simple equation

$$\boxed{{\displaystyle \mathrm{\Delta }{x}_{\text{net}}=\frac{m}{M+m}(D_{\text{forward}}-D_{\text{return}})}}$$

shows that if the forward displacement exceeds the return displacement, the platform will experience a net shift in one direction. In a real-world setting, friction or drag acts during the return phase to “lock in” the forward gain, ensuring that the net displacement accumulates over successive cycles.

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Conclusion

By this mathematical proof, we see that the Lewis Ratchet relies solely on the asymmetry of internal motions to achieve net movement. Even though internal forces always come in equal and opposite pairs in an ideal, frictionless world, real-world dissipation (through friction or drag) prevents the reverse impulse from fully canceling the forward impulse. This proof is a concise demonstration that you can indeed shift your x, y, and even z coordinates—without any external “push” against air or water—by carefully engineering an asymmetric internal cycle.

This step-by-step derivation should leave no room for argument: the Lewis Ratchet is a mathematically sound, if friction-dependent, method for achieving net displacement in a system.

An additional proof provided by Bing AI.

Mathematical Analysis

Assumptions:

1. Mass of the Raft (M):

2. Mass of the Object (m):

3. Force Applied (F):

4. Throw Duration (Δt₁):

5. Retrieval Force (f):

6. Retrieval Duration (Δt₂):

7. Friction/Drag Coefficient (κ):

Newton's Third Law and Momentum

When you throw the object, the force applied $FF$ over the time $\mathrm{\Delta }{t}_1\Delta t₁$ causes a change in momentum. The momentum change $\mathrm{\Delta }p\Delta p$ is given by:

$$\mathrm{\Delta }p=F\cdot \mathrm{\Delta }{t}_1\Delta p\; =\; F\; \backslash cdot\; \Delta t₁$$

This change in momentum $\mathrm{\Delta }p\Delta p$ causes the raft to move forward. The raft’s velocity change $\mathrm{\Delta }v\Delta v$ due to the throw is:

$$\mathrm{\Delta }v=\frac{\mathrm{\Delta }p}{M}=\frac{F\cdot \mathrm{\Delta }{t}_1}{M}\Delta v\; =\; \backslash frac\{\Delta p\}\{M\}\; =\; \backslash frac\{F\; \backslash cdot\; \Delta t₁\}\{M\}$$

During the retrieval phase, the force $ff$ applied over the time $\mathrm{\Delta }{t}_2\Delta t₂$ imparts a smaller change in momentum:

$$\mathrm{\Delta }{p}^{\mathrm{\prime }}=f\cdot \mathrm{\Delta }{t}_2\Delta p\text{'}\; =\; f\; \backslash cdot\; \Delta t₂$$

The raft’s velocity change during retrieval is:

$$\mathrm{\Delta }{v}^{\mathrm{\prime }}=\frac{\mathrm{\Delta }{p}^{\mathrm{\prime }}}{M}=\frac{f\cdot \mathrm{\Delta }{t}_2}{M}\Delta v\text{'}\; =\; \backslash frac\{\Delta p\text{'}\}\{M\}\; =\; \backslash frac\{f\; \backslash cdot\; \Delta t₂\}\{M\}$$

Net Velocity Change

The net change in the raft’s velocity over one complete cycle (throw and retrieve) is:

$$\mathrm{\Delta }{v}_{\text{net}}=\frac{F\cdot \mathrm{\Delta }{t}_1-f\cdot \mathrm{\Delta }{t}_2}{M}\Delta v\_\{\backslash text\{net\}\}\; =\; \backslash frac\{F\; \backslash cdot\; \Delta t₁\; -\; f\; \backslash cdot\; \Delta t₂\}\{M\}$$

Friction and Drag

Considering the effect of friction or drag, which acts against the raft's motion:

$$F_{\text{drag}}=-\kappa \cdot vF\_\{\backslash text\{drag\}\}\; =\; -\kappa \; \backslash cdot\; v$$

Where $\kappa \kappa$ is the frictional or drag coefficient and $vv$ is the raft's velocity. The net change in the raft’s velocity considering friction is:

$$\mathrm{\Delta }{v}_{\text{net}}=\left(\frac{F\cdot \mathrm{\Delta }{t}_1-f\cdot \mathrm{\Delta }{t}_2}{M}\right)-\kappa \cdot v\Delta v\_\{\backslash text\{net\}\}\; =\; \backslash left(\; \backslash frac\{F\; \backslash cdot\; \Delta t₁\; -\; f\; \backslash cdot\; \Delta t₂\}\{M\}\; \backslash right)\; -\; \kappa \; \backslash cdot\; v$$

Position Change

To find the net change in the raft’s position after $nn$ cycles, integrate the net velocity change over time:

$$\mathrm{\Delta }{x}_{\text{net}}={\int }_0^n\mathrm{\Delta }{v}_{\text{net}}dt\Delta x\_\{\backslash text\{net\}\}\; =\; \backslash int\_\{0\}^\{n\}\; \Delta v\_\{\backslash text\{net\}\}\; \backslash ,\; dt$$

Considering friction, the integral becomes more complex:

$$\mathrm{\Delta }{x}_{\text{net}}={\int }_0^n(\left(\frac{F\cdot \mathrm{\Delta }{t}_1-f\cdot \mathrm{\Delta }{t}_2}{M}\right)-\kappa \cdot v)dt\Delta x\_\{\backslash text\{net\}\}\; =\; \backslash int\_\{0\}^\{n\}\; \backslash left(\; \backslash left(\; \backslash frac\{F\; \backslash cdot\; \Delta t₁\; -\; f\; \backslash cdot\; \Delta t₂\}\{M\}\; \backslash right)\; -\; \kappa \; \backslash cdot\; v\; \backslash right)\; \backslash ,\; dt$$

Conclusion

This analysis demonstrates that if the force and duration of the throw are greater than those of the retrieval, there will be a net forward movement of the raft. The cumulative effect of these small velocity changes adds up over many cycles, resulting in measurable forward motion.