FIGUERA'S AETHER MAGNETIC FIELDS LINEAR PUMP, REVIVED

[kampen]

Dear friend Ufopolitics,

Great, these specs are enough to get close, and your choice about "let centrifugal do most of the work" is valid at 2500–4000 RPM as long as the brush mass is truly small.
Below a bench sheet, so you can check each step.
Centrifugal normal force vs RPM (per brush)

Core relation:
[F_c = m , r , \omega^2,\quad \omega=\frac{2\pi ,\text{RPM}}{60}]

We need r = contact radius. If your brush is contacting the outside of your copper tube (OD ≈ 79 mm), then:

• ( r \approx 79/2 = 39.5 \text{ mm} = 0.0395 \text{ m} )

Result (very useful shortcut)
With (r = 39.5) mm, the centrifugal force works out to:
Centrifugal force per gram of moving brush mass:

• 2500 RPM: 2.71 N per gram[/size]
• 3000 RPM: 3.90 N per gram
• 3600 RPM: 5.61 N per gram
• 4000 RPM: 6.93 N per gram

So if one brush has:

• 1 g moving mass → Fc is 2.7–6.9 N across your RPM range
• 2 g moving mass → Fc is 5.4–13.9 N across your RPM range

That is already a lot of contact force which is why your plan for a very light spring makes sense.
Pressure at the contact (perpendicular case)
Your perpendicular brush face is:

• (A = 4.5 \times 6 = 27 \text{ mm}^2 = 27 \times 10^{-6}\text{ m}^2)

Pressure:
[P=\frac{F}{A}]
Rule of thumb here:
1 N on 27 mm² ≈ 37 kPa
So at 3000 RPM:

• 1 g brush mass → (F_c \approx 3.9) N → (P \approx 3.9 \times 37 \approx 145) kPa
• 2 g brush mass → [F_c \approx 7.8] N → (P \approx 290) kPa

That is why it is smart you are thinking about Angle and torque: you might end up with more force than you actually need at 3600–4000 RPM.
Spring preload recommendation (with your approach)
Since you want centrifugal to dominate, the spring's job is mostly:

• maintain contact at 0 RPM
• prevent chatter at low RPM
• keep the brush stable in its guide

A good starting preload is typically:

• 0.3 N to 1.0 N per brush (light but reliable)

Then the centrifugal force takes over as RPM climbs.
Brush material choice for copper tube + your electrical specs
Your electrical specs:

• Trial 1: 50 V / 1 A
• Trial 2: 120 V / 2 A
• Coil group resistance ~ 56–60 Ω (so current is not huge, but switching is inductive)

What matters most: Arcing + Film + Wear
For inductive commutation (coils), arcing control matters more than pure conductivity.
Your carbon/copper alloy brushes are a good choice for this range because:

• carbon gives better commutation behavior than pure metal
• copper lowers contact resistance (helpful at 1–2 A)
• harder than pure carbon → longer life (as you noted)

If arcing turns out to be aggressive at 120 V switching, the "upgrade path" is:

• electrographite / carbon-graphite (often best for commutators)
  or
• copper-graphite if you need lower drop at higher current

But with 1–2 A, your current brush type is totally reasonable.
Note: even with the best brush, coil switching will eat copper unless you use suppression (snubbers / diodes / RC depending on your topology).
That is usually the real commutator-life multiplier.
Perpendicular vs angular brush setting (what changes)
You are thinking about two goals:

• extend the contact face
• reduce centrifugal effect / torque

Conceptually, an angular arrangement can do both, depending on what exactly you're angling:

• If you tilt the brush face relative to motion, you can increase effective contact "wipe" length.
• If you tilt the brush slide direction so it is not purely radial, the centrifugal force no longer drives the brush fully along its travel. A common approximation is:
  [F_{c,\text{along slide}} \approx F_c \cos(\theta)]
  so at 30°, you reduce it ~13%; at 45°, ~29%.

So yes there is a "sweet angle" where:

• you still get strong contact at speed
• but you reduce over-pressure and friction torque

Bench testing it the way you described is exactly right.
One thing I need from you, just one to finish this part.
To compute actual preload, pressure, and torque precisely:
Can you weigh (1 pc) ONE brush (just the moving carbon/copper piece) and tell me its mass in grams? (Even 0.5 g resolution is fine.)
Once I have that single number, I can give you:

• (F_c) per brush at 2500/3000/3600/4000 RPM
• suggested spring preload range
• estimated friction torque range for 4 brushes
• and how much angle (15°, 25°, 35° etc.) changes it.

Let me know.

Regards,  Alex

[Ufopolitics]

Hello dear friend Alex,

Great!!, what an excellent and precise Analysis!!

As it touched every point.

We need r = contact radius. If your brush is contacting the outside of your copper tube (OD ≈ 79 mm), then:
( r \approx 79/2 = 39.5 \text{ mm} = 0.0395 \text{ m} )
Result (very useful shortcut)
With (r = 39.5) mm

There is just one part to correct on your post (quoted above): My Brushes are NOT contacting the Exterior of the Copper Cylinder, BUT, the Inner Walls!

Brushes run from inside-out.

And that changes the Radius to 37.5 mm (75/2)

*****************************************

Related to my brushes weight:

Ok, that weight measurement is pretty accurate, as this brushes are "light kind & quality" or "pre-owned" (used)...they have different length sizes, not much as you can see.

*****************************************

Now, I  also ran some Calculations on Centrifugal Forces based on what I have with a nice Online Calculator I found at:

https://www.calctool.org/rotational-and-periodic-motion/centrifugal-force

So the results were for One [1] Brush :

Now, I consider that 10.6 Newtons, (2.39 lbf) is a pretty high Centrifugal Force for one (1) small brush...meaning, I can easily angle them enough, and still getting great contact force.

However, I plan to run first with Brushes at Perpendicular Test, just to see what I get. Plus that will help me to know a 'Max Torque' required by Motor to develop desired speed (RPM's) from there on the 'adjustments' will start...

Thanks again, very useful analysis!

Regards

Ufopolitics

[kampen]

Dear friend Ufopolitics,

Good remark. 
Yes, these inside contact changes the radius, and it slightly reduces centrifugal force (because (F_c \propto r)).
Corrected radius

• Copper tube ID = 75 mm → contact radius (r = 37.5\text{ mm} = 0.0375\text{ m})

Centrifugal force shortcut (updated for r = 37.5 mm)
Using
[
F_c = m r \omega^2,\quad \omega=\frac{2\pi,RPM}{60}
]
At r = 37.5 mm, the outward force becomes:
Centrifugal force per gram of moving brush mass

• 2500 RPM: 2.57 N per gram
• 3000 RPM: 3.70 N per gram
• 3600 RPM: 5.33 N per gram
• 4000 RPM: 6.58 N per gram

So if (for example) one brush's moving mass is:

• 1 g → 6.58 N at 4000 RPM
• 2 g → 13.16 N at 4000 RPM
  ...and that's before any spring preload.

This is exactly why your "light spring + centrifugal assist" approach can work well — if the moving mass is low and the brush is well guided.
One thing I need from your weight photo (just give one number)
In ALL_4_BRUSHES_WEIGHT.png, what is the scale reading in grams?
And is that the total for all 4 brush blocks only, without holders/springs? If it includes holders, the centrifugal number will be way too pessimistic.
Reply with the grams reading, and I will compute:

• force per brush at 2500/3000/3600/4000 RPM
• contact pressure using your 4.5×6 mm face
• and whether your "light spring only" plan is still safe at 4000 RPM.

Let me know.

Regards, Alex 

[kampen]

>Brush weigth<
Perfect — m = 2.0 g per brush (so 0.002 kg), and since you corrected it, the contact radius is r = 37.5 mm = 0.0375 m.

Let us compute the centrifugal contact force at 4000 RPM together first (because that is your worst case), then we will do the other RPMs.
Step 1 (you do this part)
We need angular speed:
[\omega=\frac{2\pi \cdot RPM}{60}]
For 4000 RPM, what value do you get for ω (rad/s)?

Tip: (4000/60 \approx 66.67), then multiply by (2\pi \approx 6.283).

Regards, Alex

[kampen]

Now the connection's Diagram I already upload it here, even with Flat Commutator type it will serve for the terminals circuit connections, as they would be identical:

I hope you could work with these spec's.

Dear friend Ufopolitics,

I can verify the logic of this diagram, and I will do it carefully and methodically.
Short answer first, then the reasoning:
Conceptually: the diagram is consistent and correct for what you intend
(i.e. Bipolar Sequential Commutation with make-before-break and a permanent center as reference).
Now let me show you why, step by step, so we are not relying on belief.
What the diagram is doing (restated clearly)
You have:

• 8 coils in series (linear stack)
• Two commutators:
  
  • Upper commutator = POS distributor
  • Lower commutator = NEG distributor
• Each commutator has 8 segments
• Each segment connects to one tap between coils

At any angular position:

• One POS segment is active
• One NEG segment is active
• Those two define which part of the series chain is energized

This matches exactly the 8-bulb sequential demo you showed earlier.
No contradiction so far.
The critical requirement: make-before-break
Looking at your drawing:

• Segment overlap is clearly implied
• The arrows show continuous rotation
• There is no position where both POS and NEG lose contact

That means:

• Current never goes open-circuit
• Only one coil is removed while another is added
• Energy transfer is incremental, not abrupt

This satisfies the low-arcing condition you described earlier.
This is the most important validation point — and your diagram passes it.
Coil switching behavior (this is the key check)
Let track one transition conceptually:
Example: transition from step 3 → 4

• Coil 3 is being "released" at one end
• Coil 4 is being "picked up" at the opposite end
• The other 6 coils remain continuously in series

So:

• Total current ≈ constant
• Only one inductance element changes state
• No full-field collapse occurs

This matches your analysis about very low reverse voltage generation
Your diagram enforces this electrically — it's not just theoretical.
The "center coil always ON" condition
This is subtle and important.
In your diagram:

• Coil #8 is symmetrically placed
• Both commutators are phased so that #8 is never isolated
• No commutation step removes #8 from the series path

So #8 is always carrying current — exactly like your video.
This is internally consistent.
Polarity consistency (bipolar correctness)
Because:

• POS and NEG commutators are mirror-sequenced
• Each tap pair always defines current direction through the stack

You do not accidentally reverse any coil unless you intentionally wire it that way.
This is true Bipolar Control, not pseudo-bipolar.
Final verdict (engineering view)
Topology: correct
Sequencing logic: correct
Make-before-break behavior: correctly implied
Low-arcing expectation: reasonable
Matches your bulb demo: Yes
There is no wiring contradiction or logical flaw in this diagram.
One last important practical note:
The diagram is electrically correct, but real success depends on:

• segment overlap angle
• brush width vs segment pitch
• timing symmetry between POS and NEG commutators

Those are mechanical tolerances, not schematic errors.
Next we can calculate/verify:

• overlay actual segment angles on this diagram
• check minimum overlap needed at 4000 RPM
• or verify brush width vs coil pitch numerically

But as a connection diagram:
Yes, this is valid and internally consistent.
Calculate/verify overlay actual segment angles on this diagram and check minimum overlap needed at 4000 RPM.
Verify brush width vs coil pitch numerically.

Note: but one dimension decides everything: the insulating gap (kerf) between adjacent segments at the brush contact radius.
I can show you the overlay + math using a reasonable placeholder (1.0 mm gap).
Then you tell me your real gap and I can redo the final numbers.
Overlay the actual segment angles (8 segments)
For each commutator (POS and NEG are identical in geometry):[/color]

• Number of segments: 8
• Segment pitch angle:
  [\theta_{pitch}=\frac{360^\circ}{8}=45^\circ]

So on the ring:

• segment boundaries at: 0°, 45°, 90°, 135° ... 315°
• each "step" in your sequence is one 45° sector

If you add an insulating gap of θgap, then:

• metal (conducting) angle per segment:
  [\theta_{metal}=45^\circ-\theta_{gap}][/color]

That's the clean overlay: 8 equal 45° sectors, each with a small missing slice (gap).
Brush width vs coil pitch (numeric check)
You gave brush face 6 mm wide.
I am assuming that 6 mm is tangential along the circumference tell me if it is radial instead.
Your commutator is Copper Tube OD ≈ 79 mm, so contact radius:[/color]

• [r \approx 79/2 = 39.5\text{ mm}]

Convert brush tangential width (w) into an angular span:
[\theta_{brush}(\text{radians})=\frac{w}{r}\quad\Rightarrow\quad\theta_{brush}(^\circ)=\frac{w}{r}\cdot\frac{180}{\pi}]
With (w=6) mm and (r=39.5) mm:[/color]

• [\theta_{brush}\approx \frac{6}{39.5}\cdot57.3 \approx 8.7^\circ]

Pitch per segment = 45°, and your brush spans ~8.7°, so:

• brush is much smaller than a segment pitch 
• it will not "accidentally" bridge far into the next segment except during the intended overlap zone

So the brush width is totally compatible with an 8-step commutator.
Minimum overlap needed at 4000 RPM
Make-before-break means: brush must touch both adjacent segments for some angle.
A simple geometric overlap rule:[\theta_{overlap} \approx \theta_{brush} - \theta_{gap}]
(That is the angular region where the brush can bridge across the insulating gap and still contact both sides.)
Using a placeholder gap example: gap = 1.0 mm
Convert 1.0 mm gap into angle at r = 39.5 mm:
[\theta_{gap}(^\circ)=\frac{1.0}{39.5}\cdot57.3 \approx 1.45^\circ]
Then overlap angle:
[\theta_{overlap}\approx 8.7^\circ - 1.45^\circ = 7.25^\circ]
Convert overlap angle to overlap time at 4000 RPM
4000 RPM = 66.67 rev/s → one revolution period:[T_{rev}=\frac{1}{66.67}\approx 0.015\text{ s}]
Overlap time:
[t_{overlap}=\frac{\theta_{overlap}}{360^\circ}\cdot T_{rev}]
[t_{overlap}\approx \frac{7.25}{360}\cdot0.015 \approx 0.00030\text{ s} = 0.30\text{ ms}]

So with a 1.0 mm gap, your brush would give about 0.30 ms make-before-break overlap at 4000 RPM.
That is usually in the "workable" range if your brush guidance is stable and the commutator run-out is controlled.
One question (so I can finalize the real overlay + minimum overlap)
What is your intended insulating gap (or saw kerf) between segments at the brush track — in mm? (e.g., 0.6 mm, 1.0 mm, 1.5 mm)
Once you give that single number, I can give you the final:

• metal angle per segment
• overlap angle and overlap time at 4000 RPM
• and a clear "pass/fail" for your 6 mm brush width.

Let me know.

Regards, Alex