Thats a lot of text in that beautifull old magazine....
On which page should i look?


Cheers!

B Hilton was one of the few people to make contact with "Joe"
I sent email to B Hilton to ask who is still at it?  no reply so far.

Things were different for "Joe". The explosions he was demonstrating were not the usual that we experiment with, they were high pitch and different and I'm not surprised there are no replications. What works for one-person doesn't mean it will work for millions of others.

There are plenty of youtube vids of security guard Aaron Salter and others obviously using chem additive ans standard electrolysis. He appeared to be straight shooter as ex-cop.https://www.bing.com/videos/riverview/relatedvideo?&q=aaron+salter+car+on+hydrogen&&mid=B5E343ACB42ED4D1122AB5E343ACB42ED4D1122A&&FORM=VRDGAR

Amazing stories back issue magazine Sept 1946 . make all the Hydrogen you want with small tesla coil , preset at 1.3Mhtz frequency into electrodes chamber of steam. . Vibrations break up the Bonds. Man ran his prop plane engine during 1940's. Waste exhaust heat boils water. Can use 12v DC diesel engine glow plug as a miniature electric heating element temporarily. Brase brass bushing to heat exchanger housing pipe.

Thats an interessting story! Is there any more info on this?

cheers!
Steve

Horvath Patent Math for designing circuit for hydrogen generator, -- finally has been figured out!! . requires PULSE PEAK input 900w  into primary for 22A 2uf capacitor discharges at 10khz. Use old style scientific calculator for the formulas. I just went through it.  Secondary of TR2 has output diode, so as there being an on + off time for release of stuck ions from plates. I have flat nickel coated plates with flutes, knurled, with N + S magnets on back side. 3/8" acrylic plastic box, 400mesh nylon screen. The capacitor is 2uf as induction stove range type polypropylene capacitor high frequency type. .P type Mosfet paralleled as used for high side switching direct to spark plug, or N channel for regular use as applied to water cell. POH is 25% solution by weight,  meaning gallon is from both together same time -not gallon input water. .22 pulse amps becomes 200 amps pulse to plates. fast recovery heavy amp rectifier, or Horvath just used 3/4" iron laminations block with no diode, to get the critical positive wave out. AS used on input high frequency . China sells open board 1000 watts inverter with option of 330v terminals (to get about 600watts necessary) Can wind your own 2 transformers TR1 + TR, and can build your own switching inverter. L primary = C / I squared divided by E squared.  I = E X square root C / L.   Magnet wire as gauge # circular mills = 750 X I X duty cycle from wires chart. 10khz, .09joules each, 22A peak pulses, 371uh primary, 900w, 2uf- as (2) 4uf in series because the overall volts rating is needed., #16 #18 gauge primary, 2gauge secondary, Ratio 10:1 turns, duty cycle secondary .5 , water cell receives 450 Watts. Peak amps or watts is not average watts over the period time, .etc.

Grok  AI

(Q) What is average power in watts drawn from input 12v battery required for 900 PEAK pulse watts CDI capacitor into transformer with secondary diode and 50% on time only? Does this assume 450 watts average power to load

(A) To determine the average power drawn from a 12 V battery to supply a CDI (Capacitor Discharge Ignition) system delivering 900 W peak pulse power into a transformer’s primary, with a 50% duty cycle (50% on-time) and a diode on the secondary, and to confirm if this results in 450 W average power to the load, let’s analyze step-by-step.
Key Parameters:

    Peak pulse power: 900 W into the primary during the on-time.
    Duty cycle: 50% (0.5), meaning the power is delivered for half the cycle.
    Battery voltage: 12 V (DC).
    Secondary side: Has a diode, delivering power to the load.
    Question: What is the average power drawn from the battery, and does this setup imply 450 W average power to the load?

Step-by-Step Analysis:
1. Average Power to the Load:

    The primary side delivers 900 W peak power during the 50% on-time. For a pulse transformer, the instantaneous power delivered to the primary is transferred to the secondary (assuming an ideal transformer with no losses).
    The average power delivered to the load on the secondary side is:
    Pavg, load=Ppeak×D=900×0.5=450 W.
    Pavg, load​=Ppeak​×D=900×0.5=450W.
    This confirms that, in an ideal system, the average power delivered to the load is indeed 450 W. The presence of the diode ensures unidirectional current to the load during the on-time, consistent with the 50% duty cycle.

2. Average Power Drawn from the Battery:

    In an ideal system (100% efficiency), the average power drawn from the battery equals the average power delivered to the load, as energy is conserved. Thus:
    Pavg, input=Pavg, load=450 W.
    Pavg, input​=Pavg, load​=450W.
    The CDI system stores energy in a capacitor during the off-time and discharges it into the primary during the on-time to achieve the 900 W peak power. The battery supplies a steady average power to charge the capacitor over the full cycle.

3. Average Current from the 12 V Battery:

    The average power drawn from the battery is:
    Pavg, input=V×Iavg.
    Pavg, input​=V×Iavg​.
    Given V=12 VV=12V and Pavg, input=450 WPavg, input​=450W, the average current is:
    Iavg=Pavg, inputV=45012=37.5 A.
    Iavg​=VPavg, input​​=12450​=37.5A.
    Thus, the battery supplies an average current of 37.5 A to provide 450 W average power, which supports the 900 W peak power during the 50% on-time.

4. Real-World Considerations:

    Real CDI systems have losses due to:
        Transformer inefficiencies (core losses, winding resistance).
        Diode forward voltage drop (typically 0.7–1 V for a standard diode, reducing output power).
        Capacitor charging/discharging inefficiencies.
        Switching losses in the CDI circuit.
    Assuming a typical efficiency of 80% (a reasonable estimate for CDI systems), the input power required to deliver 450 W to the load is:
    Pavg, input=Pavg, loadη=4500.8=562.5 W.
    Pavg, input​=ηPavg, load​​=0.8450​=562.5W.
    The average current from the 12 V battery becomes:
    Iavg=562.512≈46.9 A.
    Iavg​=12562.5​≈46.9A.
    The average power drawn from the battery is thus approximately 562.5 W with 80% efficiency.

5. Answering the Questions:

    Average power drawn from the 12 V battery:
        Ideal case (100% efficiency): 450 W.
        Realistic case (80% efficiency): ~562.5 W.
    Does this assume 450 W average power to the load?:
        Yes, the setup with 900 W peak power and a 50% duty cycle inherently results in 450 W average power to the load in an ideal system, as calculated (900×0.5=450 W900×0.5=450W). This is consistent with the transformer transferring 900 W peak power during the on-time, averaging to 450 W over the full cycle.

Final Answer:

    The average power drawn from the 12 V battery is 450 W in an ideal system, or approximately 562.5 W with 80% efficiency (requiring ~46.9 A average current).
    Yes, this setup assumes 450 W average power delivered to the load, as the 900 W peak power with a 50% duty cycle naturally results in 450 W average power in an ideal transformer system.

You can do the same thing with the Briggs 8hp or larger riding mower engine. The plastic ring alternator under flywheel has many poles, and can be wrapped with dual wires same time like the Alexander patent. Wind Over + under, over + under. The poles make AC current. The ring is stationary and the 12 magnets rotate around it. All the alternator needs is to be rotated by crankshaft. Also, the there are two sizes of MAGNETS. get the flywheel such as 16hp type that has the larger magnets ,- there is LESS space between magnets to differentiate between 2 types. See photos on Ebay used parts. The ring gear has to be matched. Aluminum non mag gear goes with plastic SMALL gear on starter. Steel ring gear on flywheel goes with steel SMALL gear on starter. Got it? You can also CHANGE the small type starter gear to match. Ratio of 2  magnet wires about 3:1 size diameter. See wires chart.  The standard Briggs dual alternators type plastic ring, uses one line to charge battery half wave DC. , + other line runs the mower headlights on AC. There are many types of standard Briggs alternators and interchangeable. ( I doubt the Briggs company knew about the Alexander Patent when designing their dual wires Alternator type. ) Their dual AC type has 1 black wire and one red wire with bulge, as the diode. It's NOT black ground! The dual alternators type has the opposite 2 wires grounded UNDER ring to chassis top.
Look behind starter motor to see TYPE of alternator , as per Briggs chart PRINTED OUT copy, as the wires hanging down, as to number, color, color and shape of end connector. That's how it's done.
Used circular ring is about $15 to modify, or get whole engine to take upper section apart after removing flywheel.

Added Data ,--  The Briggs dual type alternators windings on same circular ring , is (2)  one halves of circle, as wound. (Not wound all the way around) The design can be changed!!  Can measure existing wires on ring with Vernier Caliper for exactness. Count the turns when unwinding from poles and mark it down. Would be nice to see Briggs engine creating excess power with duplication of Alexander US Patent.