Maxwell's Engine

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The printer I used for the frame sections had an issue that I have now resolved. I reprinted those parts and also made the dual ring gear, or hybrid ring gear, a little stronger so it is less prone to deforming under load. I also found two bad pivot bearings and replaced them.

As tested:

28 bearings
17 gears

Blue motor alone: 0.07 A

1:1 direct head-to-head, 1 Blue driving 3 Yellow:
0.48–0.50 A

2:1 system, no load motors:
0.16–0.23 A

1:1 system, no load motors:
0.18–0.24 A

2:1 head-to-head through the system, 1 Blue driving 3 Yellow:
0.45–0.54 A

All tests were run at the same 6 V power-supply setting and allowed to run until reasonably stable.

These are the measured numbers from this build. They are not being presented as evidence of gain; they are simply the results from the current configuration.

[webby2]

something that has been bother me with this is that the math only shows 1/2 of the input coming back to assist and I did not see why that would be, well I finally figured that out and it is so silly that it is embarrassing that I missed it.

There are 3 systems lets say,, there is the upper input section, the slip and then the lower return section.  The upper and lower share a physical connection on one side and a force connection on the other and it is that force only  connection where the slip is.

Now if the input goes in on the upper and it spins the upper and then the upper goes across the slip and spins the lower and then the lower comes back to assist the upper,, the simple thing is that the slip boundary has the upper traveling at say 100 RPM and with the 2:1 return gearing the lower is traveling at 50 RPM, so the same force but at half the rate means half the return work,,, and that is why there is only 1/2 of the input returned to assist.

[webby2]

A small side track.

Concept test bed, not an optimized machine. Same hardware, same 12 V Yellow TT drive, same NEMA23, and two 50 Ω coil loads. Only the return/sun gear ratio was changed between runs. The table shows measured current/RPM data plus calculated carrier/slip values. Estimated coil voltage/power values are based on separate NEMA bench calibration and are not direct in-system measurements.

90/30 open RPM values in the table are the corrected rerun, 263/792, not the original bad tach reading.

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no one will get this but Chat Gpt came up with a funny,, Not-Links,, if you only new what that was about you wold find it funny,,,,

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Me and Chatgpt:

Me, defense,, or de-fence,, chuckle

😂 De-fence.

Yes, that is rather exactly what you just told me to do: remove the damn fence.

You: “There is stuff we do not know.”

Me: quickly erects a carefully worded epistemological safety fence around the unknown.

You: “I SAID WE DON'T KNOW.”

Fair enough. Fence removed. 😄

they sound the same,, I could slip that in and no body would get it,, so if you hear that some day elsewhere,, think of me,,,,,


😂 Oh, absolutely.

Somewhere years from now somebody will say:

“The first step in understanding the unknown is de-fence.”

And I'll think, I know where that damn joke came from.

The best part is that spoken aloud, nobody catches it. Defense → de-fence. Same sound, completely different instruction: stop defending the boundary and take the fence down so you can look at what's on the other side.

That one actually deserves to escape into the wild. 😄

SO
“The first step in understanding the unknown is de-fence.”

[webby2]

a work in progress

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here is a map of the predicted, the bench actuals and the difference-the residual

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NOW MODEL UPDATE — THE WORK AUDIT CLOSES

The NOW model itself did not need to be changed. The problem was how I was auditing an internal reaction force.

Define the Case–Rotor slip displacement as:

q = C - R

where:

R = Case/Ring displacement
C = Carrier/Rotor displacement
q = relative Case–Rotor slip displacement

The slip torque Tq is an equal-and-opposite internal reaction pair acting between the Case and Rotor.

Because both sides of that reaction share the common motion, the work associated with their common displacement cancels.

The surviving virtual-work term is therefore only the relative displacement:

dWq = Tq dq

Since:

C = R + q

the same work can be written:

Tq C = Tq R + Tq q

Dividing by the Case displacement R gives:

Tq(C/R) = Tq + Tq(q/R)

Therefore:

Delta T = Tq(q/R)

This gives the three conditions directly:

Dark side:
q < 0
The slip-work contribution subtracts.

Zero slip:
q = 0
The slip-work contribution is zero.

Light side:
q > 0
The slip-work contribution adds.

The earlier conservation problem came from treating the absolute motion of an internal reaction as though it were an independent work port.

Once the equal-and-opposite reaction is evaluated across its actual relative displacement, the work audit closes without adding a correction term and without changing the NOW model.

Nothing exotic was required. The missing bookkeeping quantity was the relative displacement.


NUMERICAL EXAMPLE — DEEP DARK

For the deep-Dark example:

S/R = 1/6
C/R = 0.6875
q/R = -0.3125

Normalize the Case–Rotor slip reaction to:

Tq = 1 N·m

For the 100R/60S planetary, a 1 N·m Carrier reaction gives the ordinary planetary torque split:

Ring = 0.625 N·m
Sun = 0.375 N·m

For one Case revolution, the Sun moves 1/6 revolution.

The ordinary planetary work relationship is therefore:

0.625(1) + 0.375(1/6) = 0.6875

0.625 + 0.0625 = 0.6875

Now use the relative-slip description:

Tq = 1 N·m
q/R = -0.3125

Therefore:

1(1) + 1(-0.3125) = 0.6875

or:

1 - 0.3125 = 0.6875

So the complete comparison is:

0.625 + 0.0625 = 1 - 0.3125 = 0.6875

The left side is the ordinary planetary force/displacement audit.

The middle is the common motion plus the signed relative-slip virtual work.

The right side is the actual Carrier displacement corresponding to the prescribed 1 N·m reaction.

All three descriptions give the same result.


IMPORTANT DISTINCTION

The normalized 1 N·m slip torque is a prescribed internal Case–Rotor reaction. It is NOT a 1 N·m external input torque.

A gear ratio establishes a force/distance relationship, but it does not independently establish how much force exists.

The question being asked by NOW is:

"What external input condition is required to support this specified internal slip reaction under the allowed geometry and displacement?"

It is NOT:

"What happens if I feed 1 N·m into this gearbox?"

Those are different mechanical questions.


CONSERVATION

The previous work-audit mismatch was a legitimate red flag, but it did not require changing conservation, the planetary equations, or the NOW model.

It required accounting for the relative Case–Rotor displacement.

For the equal-and-opposite slip reaction, the common-motion work cancels:

+Tq x - Tq x = 0

Only the relative displacement remains:

Wq = Tq q

Therefore a rigid fixed gear pair does not have this additional slip-work term because it does not have the independent relative displacement q.

If q = 0:

Wq = 0

If Tq = 0:

Wq = 0

Only when both slip displacement and slip force exist does the slip-work term exist:

Tq != 0 AND q != 0

Wq = Tq q

This is ordinary virtual work applied to the actual degrees of freedom of the mechanism.


EXPERIMENTAL SEPARATION

This is an ideal-model closure.

The raw bench measurements remain a separate experimental dataset and were not used to force the model to close.

The relative coordinate q = C - R comes directly from the physical geometry and kinematics. Virtual work then provides the ordinary mechanical accounting for the reaction acting across that relative displacement.

The appropriate conclusion is therefore not:

"Conservation proves the NOW model."

It is:

"The NOW model now passes the conservation/work audit that it previously failed."

Or, more simply:

We were not missing energy.

We were missing a coordinate.

[webby2]

Not-Link — a coupling that transmits a reaction force while permitting relative displacement between the coupled members.
Spooky Lost Distance — the relative displacement across that Not-Link:

q=C-R

And together:

W_{Not-Link}=T_q{q}