Babcock-Paul: Primer
Project: US Patent 8,860,273 B2, "External Field Interaction Motor". Inventors Paul M. Babcock, David J. Babcock and Phillip N. Smith; assignee Flyback Energy, Inc., Spokane WA. Filed 28 Dec 2010, granted 14 Oct 2014 [read: page image p.1].
Source gap (repeated from Step 1). The project is one 17-page patent PDF with no text layer. Page 1 was read as an image. Pages 10-17 (description and the 20 claims) were read by OCR of page images [read: page image p.10-17]; for this primer I re-ran the OCR and checked every patent quotation below against it. The seven drawing sheets were not studied in detail, so the switching circuit (FIG. 12) is known only from its written description. The patent contains no measurements, no efficiency figures, no torque curves and no equations. Everything quantitative in this primer is textbook physics applied to assumed, illustrative numbers. No conclusion about how well the real motor performs can be drawn from the documents; there is no data. Still to obtain: dynamometer data, Flyback Energy test reports, provisional application 61/290,259, publication US 2011/0156522 A1, and the incorporated patent US 7,602,157.
Source tags: [read] = from the patent; [web] = Internet source with URL; [memory] = from general knowledge, with confidence.
1. The idea in one paragraph
An ordinary electric motor pushes magnets past coils across a very narrow air gap. Babcock's motor instead lines up many small, separately switched coils along the path, and gets its push from the magnetic fields that bulge out around the ends of the moving magnet assembly, where they meet the fields of the neighbouring coils. Any coil the magnet is directly over is switched off completely, so no current flows in it and it cannot brake the magnet. Each time a coil is switched off, the energy stored in its magnetic field (the "flyback" spike that makes sparks at a switch) is caught in a capacitor and reused instead of being wasted as heat. The patent says this gives a light, cheap motor that keeps strong thrust at high speed and is very efficient. It does not claim to put out more energy than it takes in. It is an engineering idea built on ordinary physics; the open question is how well it works, and nobody has published a measurement.
2. What you need to know first
Magnetic force and field energy. Magnets and current-carrying coils push and pull on each other. A useful way to compute the force is through stored energy: a magnetic field holds energy, about B²/(2μ₀) joules per cubic metre (B is the magnetic field in tesla, T; μ₀ = 4π×10⁻⁷ T·m/A is a constant). Things move in the direction that changes this energy, so force F = −dW/dx, where W is the field energy and x the position (with the source currents held fixed the sign flips; a textbook explains why). This is why the patent says the armature is pulled along where the fields "interact to cancel, or weaken, the energy stored in the magnetic field" [read: page image p.15]. Read: Griffiths, Introduction to Electrodynamics, chapters on magnetostatics and on electrodynamics (ch. 7 covers Faraday's law, inductance and magnetic energy).
Faraday's law and back-emf. A changing magnetic flux Φ (field times area, in webers, Wb) through a coil of N turns induces a voltage ("electromotive force", emf) e = −N dΦ/dt. The minus sign is Lenz's law: the induced voltage opposes the change that made it. In a motor, the moving magnet sweeps flux through the coils and induces a voltage that opposes the supply. This is called back-emf or counter-emf (CEMF); the patent calls it "counter-motive force" or "Reverse EMF". Read: Griffiths ch. 7.
Inductors and stored energy. A coil has inductance L (henries, H): voltage needed to change its current is V = L dI/dt. A coil carrying current I stores energy E = ½ L I². That energy cannot vanish when you open the switch: the coil drives its current onward and the voltage rises until the current finds a path (a spark, a diode, a capacitor). This is the inductive flyback (or "kick"). With resistance R (ohms, Ω) in series, current changes with time constant τ = L/R. Read: Griffiths ch. 7; any introductory circuits text on RL circuits.
How motors turn electricity into motion. In a coil carrying current i against back-emf e, the electrical power converted to mechanical power is e·i (watts). Mechanically, power is force times speed, F·v, or torque times angular speed, T·ω (ω in radians per second). Energy conservation for the whole motor: P_in = F·v + losses. Read: Hughes and Drury, Electric Motors and Drives (very readable, written for non-specialists); Chapman, Electric Machinery Fundamentals.
Switching power electronics. Modern motors are driven by transistors that switch coils fully on or fully off thousands of times a second. Pulse-width modulation (PWM) sets the average current by varying the on-time. Diodes and capacitors handle the flyback energy; circuits that store it and return it rather than burn it are called energy-recovery (regenerative) snubbers. Read: Mohan, Undeland and Robbins, Power Electronics: Converters, Applications and Design.
3. The textbook view
3.1 Where motor force comes from
The force on a magnet near a coil is the gradient of field energy, or equivalently the "magnetic pressure" B²/(2μ₀) acting on surfaces where field lines bunch up. Scale check: with B = 0.5 T,
B²/(2μ₀) = 0.25 / (2 × 1.2566×10⁻⁶) ≈ 9.9×10⁴ Pa.
Over 10 cm² (10⁻³ m²) that is about 99 N, an upper bound if the field were fully used. Real motors concentrate field in a narrow air gap between iron parts because iron carries flux easily (high permeability μ), so a small magnet or current produces a large B exactly where the force is made. Field that leaks around the ends ("fringe field") is weaker and spreads out, so it usually gives less force per kilogram [memory, high confidence: standard machine-design knowledge].
3.2 Back-emf is not a loss
For one coil, Kirchhoff's voltage law reads
V = L di/dt + R i + e
where V is the applied voltage, i the current, R the resistance and e the back-emf. Multiply by i:
V i = d/dt(½ L i²) + R i² + e i.
Read term by term: supply power = rate of storing field energy + heat + power converted to motion. The e·i term is the motor's useful output. A coil with zero back-emf converts nothing. Back-emf rises in proportion to speed (e = k v for some machine constant k), so at high speed it approaches the supply voltage and current can no longer be pushed in: this sets a motor's top speed for a given supply ("voltage-limited speed").
3.3 Standard tricks already in use
- Phase switching / commutation: brushless DC and switched-reluctance motors energise each coil only when it can make useful force and leave it off otherwise [memory, high].
- Field weakening: reduce effective back-emf at high speed to extend the speed range [memory, high].
- Energy recovery: in an ordinary transistor H-bridge drive, the coil's flyback current flows back through the transistors' diodes into the supply's capacitor, so most of ½ L I² is returned automatically; dissipative clamps (resistor or Zener) are used in cheap or simple circuits [memory, high]. The patent itself cites a 2000 conference paper on a soft-switching snubber with energy recovery, and many similar IEEE papers [read: page image p.2, References Cited].
4. Where this researcher departs
| Topic | Textbook motor | Babcock patent |
|---|---|---|
| Where force is made | Narrow air gap between iron parts; magnet sits over the coil | Arc-shaped regions outside the armature, between its leading/trailing edges and the neighbouring coils. "The increased area that the fields interact in compensates for loss of force" [read: page image p.15] |
| Coil under the magnet | Often energised (it is where the gap force is) | Held open-circuit, zero current. All three independent claims (device claim 1, method claim 5, device claim 14) recite this "substantially zero-current state" limitation, and it is the point of novelty [read: page image pp.16-17] |
| Back-emf | The necessary channel of energy conversion, e·i | To be "substantially reduced or avoided": "it cannot induce counter-motive force because those windings may be in open condition and electrical current cannot flow" [read: page image p.15] |
| Flyback energy | Returned to supply bus by the drive's diodes, or burned in a clamp | Captured in a capacitor "at a voltage above the source input voltage" [read: page image p.14, FIG. 12 text], used "to supplement the input power source to increase motor efficiency or ... to run external devices or circuits" [read: page image p.15] |
| Efficiency statements | Measured, typically stated with a test method | "A motor armature that produces little or no counter-motive force can be driven to a given amount torque and speed with much less electrical power" [read: page image p.15]; "has produced an extremely efficient and powerful D.C. electric motor" [read: page image p.16]. No numbers |
| Iron | Iron stator and armature cores carry the flux | Iron is still used. The stator core "is a continuous ring of ferrous material" (FIG. 3) [read: page image p.12], and each armature has a ring 510 "formed of a steel or ferrous material" (FIG. 5) [read: page image p.13], called a "ferrous horseshoe ring" [read: page image p.13] and a "horseshoe on horseshoe arrangement" [read: page image p.15]. The "non-metallic housing and/or reduction or elimination of metallic or high magnetic permeability shielding" [read: page image p.10] refers only to the housing and shielding |
| Impedance | Inductive reactance X = 2πfL is fixed by L and frequency f | Flyback capture "greatly reduces the input impedance of the electromagnets" [read: page image p.15] |
So the claimed novelty is where the force is made (the fringe field outside the gap), not an ironless motor.
What holds. (1) Force from fringe fields is real physics; nothing forbids it. (2) An open coil carries no current, so it exerts no braking force and takes no power. The patent wording is slightly off: a voltage is still induced across the open coil (Faraday's law does not switch off), but with no current it does nothing. (3) Capturing ½ L I² in a capacitor is standard and works.
Where the inferences go too far. - "Much less electrical power" for a given torque and speed. The open coil makes no drag, but it also makes no thrust. All the mechanical power F·v must still pass through the coils that are on, via their own e·i. So input power can never fall below F·v; the only thing that can shrink is losses. If the motor were already 90% efficient, the most any trick could save is the remaining 10%. - "Reduces input impedance." Switching does not change L. What a higher-voltage drive does is make the current rise faster (V = L dI/dt), which can look like "lower impedance" but is not. - Recovered energy as a "power source". The captured energy came from the supply a moment earlier. Returning it cuts waste; it cannot add energy.
The patent's own generator sentence. Near the end, the patent says the machines "may be operated as a generator by allowing current to flow in a coil segment while armature flux has substantial coupling to the coil", so that "Kinetic energy of the armature and any connected load (e.g., flywheel) can be recovered by storing the energy coupled into the stator coil segments in a battery" (regenerative braking) [read: page image p.16]. This is the cleanest evidence in the patent itself that back-emf times current, e·i, is the energy-conversion channel: let current flow in a coil that the magnet's flux links, and energy moves between motion and electricity. Run with the current one way it brakes and charges a battery; run the other way it drives. Holding a coil at zero current shuts that channel in both directions.
Why it sits in an alternative-science collection. The patent does not say. One clue the earlier reviews missed: its list of cited prior art includes two items by Bedini, US 6,392,370 and US 2002/0097013 [read: page image p.2]. US 6,392,370 is John C. Bedini's "Device and method of a back EMF permanent electromagnetic motor generator", whose text describes receiving "available excess energy from back EMF" [web: https://patents.google.com/patent/US6392370 ; https://www.freepatentsonline.com/6392370.html]. That patent belongs to the over-unity "back-EMF" tradition. Babcock's patent cites it but itself makes no excess-energy claim, and Flyback Energy publicly said it "never claimed to have a free energy machine" [web: https://www.spokanejournal.com/articles/9211-flyback-prepares-to-market-energy-capturing-technology]. The shared vocabulary ("reverse EMF", "flyback capture") is probably why the two get filed together.
5. Worked example: the energy ledger of one switched coil
Question: follow every joule through one pulse of one coil with flyback capture, then scale to a 1 kW motor. Does capture make the motor more efficient, and by how much? (All parameter values are my illustrative choices; the patent gives none.) The script example.py reproduces every number:
.venv/bin/python review/Babcock-Paul/example.py
Parameters. Supply V_s = 48 V. Capture capacitor held at V_c = 96 V (above the supply, as the patent describes). Coil L = 1 mH, R = 0.5 Ω, so τ = L/R = 2 ms. While the coil is active the moving magnet induces back-emf e = 24 V (taken constant). The switch opens when current reaches I_pk = 10 A.
Step 1: switch on. The circuit equation 48 = L di/dt + 0.5 i + 24 has solution
i(t) = I_∞ (1 − e^(−t/τ)), with I_∞ = (48 − 24)/0.5 = 48 A.
Time to reach 10 A: t_on = −τ ln(1 − 10/48) = 2 ms × 0.2336 = 0.467 ms.
Energy from the supply = ∫ 48 i dt = 48 × 48 × [t_on − τ(1 − e^(−t_on/τ))] = 2304 × [0.4672 − 2 × 0.2083] ms = 2304 × 0.0505 ms ≈ 116.5 mJ. Where it went:
| Destination | Energy |
|---|---|
| Mechanical work, ∫ e i dt (exactly half, since e = V_s/2) | 58.2 mJ |
| Heat in R, ∫ R i² dt | 8.2 mJ |
| Stored in field, ½ L I² = ½ × 0.001 × 100 | 50.0 mJ |
| Total | 116.5 mJ (matches supply) |
Step 2: switch off, flyback into the capacitor. The coil current now flows through a diode into the 96 V capacitor: −96 = L di/dt + 0.5 i + 24. Solution: i(t) = (I_pk + A) e^(−t/τ) − A with A = (96 + 24)/0.5 = 240 A. The current reaches zero after t_off = τ ln(250/240) = 0.082 ms. The 50 mJ stored splits into:
| Destination | Energy |
|---|---|
| Into the capacitor, ∫ V_c i dt | 38.9 mJ |
| Mechanical work (current still flows, so the coil still pushes) | 9.7 mJ |
| Heat in R | 1.3 mJ |
| Total | 50.0 mJ |
How the patent's own circuit does this (FIG. 12). The patent describes a four-step sequence per pulse [read: page image p.14]. Path A-A': the pulse starts by drawing on capacitor 1215, which holds the flyback energy stored on the previous cycle. Path B-B': once the capacitor has discharged, the source supplies the coil. Path C-C': at the end of the pulse the controller activates "a transitional voltage clamp" that carries the coil current briefly while the switches turn off (in the companion FIG. 10 this clamp is a gas tube 1015, "a transitional voltage clamp between pulse and capture circuits"; in the FIG. 12 drawing it is a switched resistor-capacitor branch) [read: page image p.14; deep read]. Path D-D': the "leakage energy remaining in the solenoid coil" is "recaptured and stored for use on the subsequent cycle" in capacitor 1215; a blocking rectifier 1205 lets that capacitor sit above the supply voltage [read: page image p.14]. The deep read found this to be a standard energy-recovery topology: an H-bridge with a dissipative snubber clamp (C) and a regenerative recovery path into a DC-link capacitor (D), close to the asymmetric half-bridge and "C-dump" converters used in switched-reluctance drives. Only the sequencing and per-segment control may be new; energetically nothing is unusual [deep read; memory, high].
Captured energy is 38.9 mJ out of 116.5 mJ drawn: about one third, and it can never exceed the ½ L I² that the supply just put in.
Step 3: scale up. Mechanical work per pulse = 58.2 + 9.7 = 68.0 mJ. For a 1000 W motor (for example F = 50 N at v = 20 m/s, since F·v = 1000 W) you need 1000 / 0.068 ≈ 14,700 pulses per second summed over all coils. About 573 W of flyback energy cycles through the capacitor. Net input power for three drive designs:
| Drive design | Net input power | Efficiency (1000 W / input) |
|---|---|---|
| a) Clamp burns all flyback energy as heat | 1714 W | 58% |
| b) Capture, 90% returned to the supply | 1198 W | 83% |
| c) Capture, 100% returned (ideal) | 1141 W | 88% |
What the numbers teach. 1. Capture helps a lot only against a wasteful baseline (case a). A normal transistor bridge drive already returns flyback energy through its diodes, so against a good conventional drive the gain is much smaller. That is consistent with Flyback Energy's own modest figures for its switch products (2-8% for lighting, 8-10% for cell-tower power conversion; company statements, no independent test reported) [web: https://www.spokanejournal.com/articles/9211-flyback-prepares-to-market-energy-capturing-technology]. 2. Even the ideal case needs more than 1000 W in, because resistance still takes 141 W. Input ≥ F·v + losses, always. 3. The open, overlapped coil (Step 4 of the script): a magnet passing over it still induces, say, 30 V across its terminals, but with i = 0 the power e·i = 0 and its force is 0. It neither brakes nor helps. All 1000 W must come through the coils that are on. 4. "Impedance" (Step 5 of the script): with L fixed at 1 mH, driving at 96 V instead of 48 V brings the current to 10 A in 0.144 ms instead of 0.467 ms. That is V = L dI/dt at work, not a change in the coil's impedance.
6. Evidence table
No experiment of any kind on this motor is on record, so "Tested, not supported" applies to nothing. Verdicts are the arbiter's.
| Claim | Tested? | Result | Replicated? | Verdict |
|---|---|---|---|---|
| 1. Thrust from fields interacting in arc-shaped regions outside the armature, not across a narrow gap | No public test | None | No | Plausible, untested (that thrust exists: Established magnetostatics; that it is competitive with gap motors: never tested) |
| 2. Overlapped coil held at zero current removes back-emf drag, allowing high thrust at speed and "much less electrical power" | No public test | None | No | Weak form (no drag current, higher voltage-limited speed): Established principle, untested in this geometry. Strong form, literal (input below F·v): Refuted by calculation. Strong form, charitable (lower losses): Not testable as stated, never tested |
| 3. Flyback energy captured in a capacitor and reused as a secondary power source | Technique: yes, widely in power electronics. This motor: no | Technique works; motor gain unmeasured | Technique: yes | Technique Established; gain in this motor Plausible, untested (a few percent likely); reading as net energy source Refuted by calculation (not claimed by the patent) |
| 4. Lighter, cheaper motor with high thrust-to-weight (non-metallic housing, little metallic or high-permeability shielding, modular coils; stator and armature still use iron cores) | No | None | No | Never properly tested (not dismissed; never reported) |
Other context: patent status Expired for non-payment of maintenance fees; a security interest was assigned to RPR Ventures LLC in Dec 2016 [web: https://patents.google.com/patent/US8860273B2/en]. Why it lapsed is unknown; there is no evidence of suppression.
7. What a learner should take away
Solid. Force from fringe fields, switching coils off when they would only brake, and capturing inductive flyback energy are all real, standard physics and engineering. The patent's qualitative force rule (attraction where field energy is reduced) is correct.
Wrong, and why the mistake is instructive. The sentence "little or no counter-motive force ... much less electrical power" treats back-emf as a drag to be eliminated. It is really the channel through which electrical energy becomes motion (power = e·i). Remove it from a coil and that coil stops doing work. This is the single most common misunderstanding in "back-EMF" motor claims, including the over-unity ones in the Bedini tradition that this patent cites. Learn to ask: where in the energy ledger does F·v come from? Likewise, "reduced impedance" confuses faster current rise (higher voltage) with a change in the coil itself.
Genuinely open. Whether this geometry gives competitive thrust per kilogram, given that fringe fields are usually weaker than gap fields; how much thrust is lost by leaving overlapped coils idle versus how much loss is saved; and what the FIG. 12 circuit and the incorporated patent US 7,602,157 achieved in measurement. A finite-element magnetic model (force versus armature position, and stray field) plus one dynamometer comparison against a conventional brushless motor of the same mass would settle all four claims. The expected result under accepted physics is an efficiency below 100% with at most a few percent gain from energy recovery.
A useful control case. This project is an honest efficiency patent that shares vocabulary with over-unity claims. Comparing it with a true excess-energy claim sharpens the reader's eye for the difference.
8. Check yourself
- A motor delivers torque T = 20 N·m at 1500 rpm. What is the minimum electrical input power, whatever back-emf scheme it uses?
- A coil with L = 2 mH carries 15 A when it is switched off. What is the most energy a flyback capture circuit could possibly recover from that event, and why can it not be more?
- A magnet sweeps past an open-circuited coil and induces 40 V across its terminals. What force does the coil exert on the magnet, and what power does it take?
- In the worked example, why does capture raise efficiency from 58% to 83% here, but would give much less benefit compared with a normal transistor bridge drive?
- Someone says, "the flyback circuit reduces the coil's impedance from 6.3 Ω to near zero at 1 kHz." What is the coil's inductive reactance at 1 kHz if L = 1 mH, and what is actually changing when a higher drive voltage makes the current rise faster?
9. Further reading
Mainstream - D. J. Griffiths, Introduction to Electrodynamics (ch. 7: Faraday's law, inductance, energy in magnetic fields). - A. Hughes and B. Drury, Electric Motors and Drives: Fundamentals, Types and Applications (Newnes): back-emf, voltage-limited speed, brushless and switched-reluctance drives. - S. J. Chapman, Electric Machinery Fundamentals (McGraw-Hill). - N. Mohan, T. Undeland and W. Robbins, Power Electronics: Converters, Applications and Design (Wiley): switching, snubbers, energy recovery. - R. Krishnan, Switched Reluctance Motor Drives (CRC Press): coils energised only when useful, and the energy-return circuits that go with them.
The researcher's own work and context - P. M. Babcock, D. J. Babcock, P. N. Smith, US 8,860,273 B2, "External Field Interaction Motor" (2014): https://patents.google.com/patent/US8860273B2/en - Related applications cited in the patent: provisional 61/290,259; US 2011/0156522 A1; incorporated US 7,602,157 ("Supply Architecture for Inductive Loads") [read: page image p.14, cited by number only; not read]. - Second incorporated reference: US 2009/0073726 (Babcock et al., "Current Waveform Construction to Generate AC Power with Low Harmonic Distortion from Localized Energy Sources"), incorporated for its AC operation (FIG. 3A) [read: page image pp.14-15, cited by number only; not read]. - Spokane Journal of Business on Flyback Energy: https://www.spokanejournal.com/articles/9211-flyback-prepares-to-market-energy-capturing-technology - For contrast, the over-unity back-EMF patent it cites: J. C. Bedini, US 6,392,370: https://patents.google.com/patent/US6392370
Answers
- ω = 1500 × 2π / 60 = 157.1 rad/s. P = T·ω = 20 × 157.1 ≈ 3.14 kW. The input must be at least 3.14 kW plus losses.
- E = ½ L I² = ½ × 0.002 × 225 = 0.225 J. The coil holds only the energy the supply put into it; some of that also goes to resistance and to mechanical work while the current decays (as in Step 2, where only 38.9 of 50 mJ reached the capacitor).
- Zero force and zero power. The voltage exists (Faraday's law), but with no closed path the current is zero, and both force and power e·i need current.
- In case (a) the baseline throws away all the flyback energy as heat, so recovering it makes a large difference. A normal bridge drive already sends flyback current back into the supply's capacitor through the transistor diodes, so the baseline is already close to case (b) or (c), and the remaining gain is small.
- X = 2πfL = 2π × 1000 × 0.001 ≈ 6.3 Ω, and it stays 6.3 Ω: switching does not change L. A higher voltage makes dI/dt = V/L larger, so the current reaches its target sooner (0.144 ms at 96 V versus 0.467 ms at 48 V in the example). The coil is the same; the drive is harder.