Energetic Forum  
Facebook Twitter Google+ Pinterest LinkedIn Delicious Digg Reddit WordPress StumbleUpon Tumblr Translate Addthis Aaron Murakami YouTube ONLY 13% OF SEATS AVAILABLE!!!*** 2017 ENERGY CONFERENCE ***


* NEW * BEDINI RPX BOOK & DVD SET: BEDINI RPX


Go Back   Energetic Forum > >
   

Renewable Energy Discussion on various alternative energy, renewable energy, & free energy technologies. Also any discussion about the environment, global warming, and other related topics are welcome here.

Bedini RPX Sideband Generator
Reply
 
Thread Tools
  #151  
Old 08-29-2017, 03:28 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Commutator and "Latching Reed Switch".

This setup is working flawlessly; Next step is to extend the commutator hub to a foot in length for the 3 trigger magnet levels with enough space in between to eliminate magnetic interference.


https://www.youtube.com/watch?v=qUqdYdArys4
__________________
 

Last edited by Allen Burgess; 08-29-2017 at 03:55 PM.
Reply With Quote

Download SOLAR SECRETS by Peter Lindemann
Free - Get it now: Solar Secrets

  #152  
Old 08-29-2017, 09:57 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Twin DC fan foot long commutator

This is my third Android Phone video, hopefully they'll improve:


https://www.youtube.com/watch?v=EJBp21wlYGk
__________________
 
Reply With Quote
  #153  
Old 08-30-2017, 01:39 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
commutator with reed switches:

Here's a still shot of the commutator with Reed Switches in position:
Attached Images
File Type: jpg WIN_20170829_183000.JPG (136.9 KB, 28 views)
__________________
 
Reply With Quote
  #154  
Old 08-30-2017, 02:35 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Geoffry S. Miller.

Geoffry S. Miller's commutator, brought over from a comment by Vidbid off the "Newman Motor Finally Explained" thread.

Apparently, my latching Reeds and DC fan commutator does the same kind of job for around $10:
Attached Images
File Type: png miller.png (584.3 KB, 26 views)
__________________
 

Last edited by Allen Burgess; 08-30-2017 at 04:14 PM.
Reply With Quote
  #155  
Old 09-01-2017, 11:01 PM
Bodkins's Avatar
Bodkins Bodkins is offline
Gold Member
 
Join Date: May 2008
Posts: 1,051
Quote:
Originally Posted by Allen Burgess View Post
This setup is working flawlessly; Next step is to extend the commutator hub to a foot in length for the 3 trigger magnet levels with enough space in between to eliminate magnetic interference.


https://www.youtube.com/watch?v=qUqdYdArys4
Great Idea well done Man!
__________________
 
Reply With Quote
  #156  
Old 09-26-2017, 03:42 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Magnets

I traveled back to Northern California and recovered 4 of the world's most powerful Neo magnets; Two 4" disks, a 4" sphere and a 4" tube. I plan to return these magnets to Costa Rica where I have a 10 Henry inductor coil with a laminated H core.

I need to equal 10 Teslas of electro magnetic magnet strength to neutralize the attraction force of the permanent magnets.

My prototype neutralization oscillator was merely a tiny fraction of the strength with only a few small button magnets on a 12 volt electro-magnet coil.

My next "Neutralization Oscillator" should generate some meaningful output.
__________________
 
Reply With Quote
  #157  
Old 09-28-2017, 06:52 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Powerful Disk Magnets.

Here's a picture of two large 4" x 1" neo disk magnets attached to a plastic spool through the center with threaded rod for safety. This picture was taken aboard my 40' Coronado in Northern California: Bound for Costa Rica!

Any ferrite "Keeper" would neutralize the permanent magnet fields, and confine the attraction to inside the loop.
Attached Images
File Type: jpg IMG_20170927_202037_BURST001_COVER.jpg (63.3 KB, 25 views)
__________________
 

Last edited by Allen Burgess; 09-28-2017 at 08:23 PM.
Reply With Quote
  #158  
Old 09-30-2017, 04:06 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Bifilar Electro-Magnet coil.

My first "Dragone Magnet Pump" consisted of a bifilar air core coil sandwiched between the two Neo disks in the picture above. It was controlled by a DPDT reed switch and D.C. battery operated magnet commutator.

The bifilar coil does not generate BEMF when current is interrupted. The output from the bifilar magnet pump was gushing and copious, with no cancellation effect. The poles of the bifilar are in the center of the hole.

Currently I am in possession of my "Quadfilar Spiral" coil. I plan to construct an oscillator with the two neo disks on one side of the coil and a ferrite rod with a strong neo in attraction for piston rod suspended overhead by strong elastic bands. The bifilar spiral should neutralize the magnet force below and free the overhead magnet rod to spring away and then attract back, without the need for an 18% cycle span to redirect the back-spike.

This advanced model bifilar neutralization oscillator should run itself silent and non polluting while generating abundant free power.
__________________
 

Last edited by Allen Burgess; 09-30-2017 at 04:11 PM.
Reply With Quote
  #159  
Old 10-02-2017, 09:02 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Costa Rica Magnets

I'm back home in Costa Rica with all the magnets I left California with. I built a loop of ceramics between the powerful neo's and no magnet force was noticeable outside the special "Sport Utility Box" I carried through the airports.

There's no longer anything holding the project back now!
__________________
 

Last edited by Allen Burgess; 10-05-2017 at 12:27 AM.
Reply With Quote
  #160  
Old 10-05-2017, 12:45 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Magnet bifilar coil blade switch and capacitor.

This photo below is of my original "Dragone Magnet Pump" bifilar air core coil, a blade switch and a 2500 volt micro-wave capacitor.

Behind on the left is my "Spiral Quadfilar":

My next two tests firstly involve pulsing the bifilar coil sandwiched between the two neo disks in attraction at neutralization power, then re-closing the blade switch to see how much power returns to the capacitor from the re-gauging of the field.

Simply shorting a capacitor can result in gain from "Ground Scouring". Energy from the Dirac sea charges the capacitor when it's charge level falls below ambience.

Secondly, I want to measure the inductance of the twin bifilar spiral to see if there is a phantom inductance from the coupling.
Attached Images
File Type: jpg IMG_20171004_182006.jpg (160.0 KB, 11 views)
__________________
 

Last edited by Allen Burgess; 10-15-2017 at 04:39 PM.
Reply With Quote
  #161  
Old 10-05-2017, 04:01 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Negative Inductance

This video shows a reading of increasing "Negative Inductance" in one serial bifilar coil in a Quadfilar of two while in self resonance. This indicates the presence of a magnetic field that is spontaneously building strength:

https://www.youtube.com/watch?v=uP2xEKH4qdc


Below is a picture of a serial bifilar air core coil sandwiched between two powerful Neo disk magnets in attraction
as the original "Dragone Magnet Pump" and Art Porter's solid state GAP:
Attached Images
File Type: jpg IMG_20171005_092513.jpg (118.8 KB, 10 views)
__________________
 

Last edited by Allen Burgess; 10-15-2017 at 11:05 PM.
Reply With Quote
  #162  
Old 10-14-2017, 10:23 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Neo disk oscillator.

This is a picture of a Neo (Ring 1/4" center hole) disk oscillator:

Look at the distance between the suspended Neo disk and the top of the bifilar neutralization coil. A sister Neo disk is positioned at the bottom of the coil in attraction: A stronger overhead spring would permit the top Neo disk to oscillate more closely to the top of the pulse coil thus increasing output.

The wall transformer is sending 14 volts to the coil. I am getting a solid "Jounce" out of the pulse. This build should generate usable free power. I'll upload a video of it oscillating soon.
Attached Images
File Type: jpg IMG_20171014_155310.jpg (68.9 KB, 26 views)
__________________
 

Last edited by Allen Burgess; 10-16-2017 at 12:10 AM.
Reply With Quote
  #163  
Old 10-16-2017, 09:41 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Neo disk oscillator.

Here's a new video of the Neo (Ring) disk oscillator with the top Neo disk closer to the top of the bifilar coil:

https://www.youtube.com/watch?v=lBtif6_XY7s

The bifilar coil and Neo disc magnet underneath are held down by a bungee cord attached to a heavy hardcover book. This is needed because the lower magnet and coil leap up off the table top and stick to the overhead Neo disk. The 4 ceramic rings on the upper Neo are threaded with copper wire and attached to a cabinet handle above by strong pony tail elastic bands. The throw is around 1/2" inch, plenty for the oscillation. In between are a few skein's of wire and some plastic spacers to prevent a collision between the powerful disk above and the coil and Neo disk below.
__________________
 

Last edited by Allen Burgess; 10-16-2017 at 10:27 PM.
Reply With Quote
  #164  
Old 10-17-2017, 12:12 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Better Neo disk oscillator video

Here's a better video:

https://youtu.be/PBOhkMouhSU
__________________
 

Last edited by Allen Burgess; 10-17-2017 at 02:00 AM.
Reply With Quote
  #165  
Old 10-17-2017, 07:41 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Measure of magnet force.

General question for the group:

The series bifilar air core coil was shop wound and coated in polyurethane. The inductance measures 75.2 mH without the magnets and 75.5 mH with the magnets attached. The input is 1500 mA at 14.5 volts for one second.

The question is, using Joseph Henry's formula for Inductance; What is the magnet force generated by the coil in Gauss for the one second?

We may need member bistander to help verify any results!

(mH is milli-henrys, and mA is milli-amps or 1000 th's of a Henry and 1000 th's of an Amp) (1 Tesla = 10,000 Gauss)

Working on the assumption that the ratio of input to output is equal to the 3.3 sheer to push pull force, we should be able to calculate the output of the oscillator in Watts!
__________________
 

Last edited by Allen Burgess; 10-18-2017 at 12:59 PM.
Reply With Quote
  #166  
Old 10-17-2017, 10:11 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
COP test.

There's a photograph of a DPDT blade switch in comment 160 above. Doubling the electrodes on the bifilar coil would allow 2 DMM's to attach to one side of the blade switch; One in series for amperage and the other in parallel for voltage. This would permit a quick and dirty COP measurement on the flip flop. I also received several timer switchs in the mail recently that will enable me to precisely channel exactly one second of pulse power to the neutralization coil.

An accurate pre-calculation would help confirm the test measurements. These results will finish my contribution to this thread. There's a solid chance we may be able to confirm with a high degree of accuracy the existence of an OU COP coupled with a solid theory.
__________________
 

Last edited by Allen Burgess; 10-17-2017 at 10:30 PM.
Reply With Quote
  #167  
Old 10-18-2017, 12:58 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Electromagnetic equation

Quote from smOky2,

"The way you would need to go about this, is to use the electromagnetic equation, with the dimensions of your coil and by the current (Amps) you can determine how many Gauss the coil would produce.

looking at it like that, ( in Teslas, which is 10^4 Gauss)

T = 4(Pi)(10^-7)*(number of turns ^2)*(Amps)/ (Length)(Area^2)"

This equation is for an air core coil.

Look how simple this is; The coil dimensions factor into units of inductance, so the answer can be solved for simply by multiplying Henries times Amps: 75.5 mH x 1500 mA = 113250 or 113.25 Gauss.
__________________
 

Last edited by Allen Burgess; 10-18-2017 at 01:19 PM.
Reply With Quote
  #168  
Old 10-18-2017, 05:59 PM
bistander bistander is online now
Silver Member
 
Join Date: Apr 2015
Posts: 912
Help verify?

Quote:
Originally Posted by Allen Burgess View Post
...
We may need member bistander to help verify any results!
...
Quote:
Originally Posted by Allen Burgess View Post
...
Look how simple this is; The coil dimensions factor into units of inductance, so the answer can be solved for simply by multiplying Henries times Amps: 75.5 mH x 1500 mA = 113250 or 113.25 Gauss.
Wrong. The product of inductance (Henries) times current (Amperes) results in flux (Webers), not flux density (Webers per square meter or Tesla or Gauss). Neither are units of force (Newtons) or power (Watts).

bi
__________________
 
Reply With Quote
  #169  
Old 10-18-2017, 07:59 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Inductance formula.

@bistander,

Thanks for responding to my inquiry:

I initially stated that an Inductor of 1 Henry of inductance would generate 1 Tesla of magnetic force with an electrical input of 1 Watt/hour.

1 Watt hour is equal to 3600 Joules.

1 Ampere is equal to 1 coloumb per second.

1 Coloumb is equal to 1 Joule/ volt.

Therefore a magnetic force of 1 Tesla would require a current of 3600 Amperes in an Inductor of 1 Henry of Inductance, right?

a Joule is equal to .7377 foot pounds, so 1 Tesla of magnetic force should lift an SUV of 4880 pounds. Correct me if I'm wrong.

What do you think the magnetic force in Gauss is in the bifilar coil of 75.5 mH
at 1500 mA?

How do you feel we should go about converting Webers of flux to Webers of flux density?

1 Weber of Flux is equal to 100,000,000 gauss per square centimeter right? What fraction of a Weber of Flux is the product 113250 of the Inductance times the Amps?

Let's say I attach a spring scale to a piece of iron and see how much weight it measures before the coil pulls away from it; Do you feel this would produce a more accurate measure of magnetic attraction strength?
__________________
 

Last edited by Allen Burgess; 10-18-2017 at 08:39 PM.
Reply With Quote
  #170  
Old 10-18-2017, 11:40 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Ferrite core.

Nevertheless, magnetic strength formula aside; I believe the bifilar coil is understrength and have ordered Iron powder and Magnetite to cold mold a high perm ferrite core for the bifilar air core coil.

@bistander,

How come smOky2's formula in comment #167 above has a T instead of a W at the beginning of his equation?
__________________
 

Last edited by Allen Burgess; 10-19-2017 at 12:46 AM.
Reply With Quote
  #171  
Old 10-19-2017, 12:39 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Electromagnet formula.

It appears member bistander's correct:

I see the Magnetic Constant (4 x PI x 10-7.) in this formula I failed to include in my calculation:

"Calculate the force by writing the equation:

F = (n x i)2 x magnetic constant x a / (2 x g2)

Where, F = force, i = current, g = length of the gap between the solenoid and a piece of metal, a = Area, n = number of turns in the solenoid, and the magnetic constant = 4 x PI x 10-7.

Analyze your electromagnet to determine its dimensions and the amount of current you will be running through it. For example, imagine you have a magnet with 1,000 turns and a cross-sectional area of 0.5 meters that you will operate with 10 amperes of current, 1.5 meters from a piece of metal. Therefore:

N = 1,000, I = 10, A = 0.5 meters, g = 1.5 m

Plug the numbers into the equation to compute the force that will act on the piece of metal.

Force = ((1,000 x 10)2 x 4 x pi x 10-7 x 0.5) / (2 x 1.52) = 14 Newtons (N)".
__________________
 

Last edited by Allen Burgess; 10-19-2017 at 12:53 AM.
Reply With Quote
  #172  
Old 10-19-2017, 01:02 AM
bistander bistander is online now
Silver Member
 
Join Date: Apr 2015
Posts: 912
Nonsense

Quote:
Originally Posted by Allen Burgess View Post
@bistander,

Thanks for responding to my inquiry:

I initially stated that an Inductor of 1 Henry of inductance would generate 1 Tesla of magnetic force with an electrical input of 1 Watt/hour.

1 Watt hour is equal to 3600 Joules.

1 Ampere is equal to 1 coloumb per second.

1 Coloumb is equal to 1 Joule/ volt.

Therefore a magnetic force of 1 Tesla would require a current of 3600 Amperes in an Inductor of 1 Henry of Inductance, right?

a Joule is equal to .7377 foot pounds, so 1 Tesla of magnetic force should lift an SUV of 4880 pounds. Correct me if I'm wrong.

What do you think the magnetic force in Gauss is in the bifilar coil of 75.5 mH
at 1500 mA?

How do you feel we should go about converting Webers of flux to Webers of flux density?

1 Weber of Flux is equal to 100,000,000 gauss per square centimeter right? What fraction of a Weber of Flux is the product 113250 of the Inductance times the Amps?

Let's say I attach a spring scale to a piece of iron and see how much weight it measures before the coil pulls away from it; Do you feel this would produce a more accurate measure of magnetic attraction strength?
Quote:
1 Watt hour is equal to 3600 Joules.

1 Ampere is equal to 1 coloumb per second.

1 Coloumb is equal to 1 Joule/ volt.
These 3 equivalencies are correct except Coulomb is misspelled. The rest of your post is rubbish. And why ask me those questions when you don't believe my answers anyway and treat me with disrespect?
__________________
 
Reply With Quote
  #173  
Old 10-19-2017, 01:17 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Respect

@bistander,

You failed to calculate the correct answer!

Watch this:

https://www.youtube.com/watch?v=wTxh6zHXxIw
__________________
 

Last edited by Allen Burgess; 10-19-2017 at 01:30 AM.
Reply With Quote
  #174  
Old 10-19-2017, 01:49 AM
bistander bistander is online now
Silver Member
 
Join Date: Apr 2015
Posts: 912
What?

Quote:
Originally Posted by Allen Burgess View Post
@bistander,

You failed to calculate the correct answer!

Watch this:

https://www.youtube.com/watch?v=wTxh6zHXxIw
I wasn't calculating anything. I don't know what I'm doing here. Good bye.
__________________
 
Reply With Quote
  #175  
Old 10-19-2017, 01:51 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
@bistander,

Henries X Amperes X Mu Naught = Magnetic Force in Newtons?

Did I supply enough data for you to calculate the correct answer for us, or are you going to ridicule me for another incorrect answer?

The missing factor is an approximation of the permeability of an air core. The Mu naught factor is practically infinitesimal, so the answer of 113 Gauss has to be very close to correct, right?

Stay out of here you turd.
__________________
 

Last edited by Allen Burgess; 10-19-2017 at 02:10 AM.
Reply With Quote
  #176  
Old 10-19-2017, 02:23 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Bistander's wrong.

I realized that bistander is wrong as usual. The inductance measurement includes the permeability of the air core, so the Mu naught factor is irrelevant.

3600 Joules (1 Watt/Hour) of input, to a coil of 1 Henry of Inductance, generates 1 Tesla of Magnetic Force; As I maintained from the outset. It's that simple!
__________________
 

Last edited by Allen Burgess; 10-19-2017 at 02:37 AM.
Reply With Quote
  #177  
Old 10-19-2017, 02:59 PM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Test confirmation.

I measured my 12 volt electro-magnet for Inductance, and was very surprised to find that it was 73.5 Henrys; very close to the Inductance of the air core bifilar. Armed with this coincidental equality, I moved forward with a comparison test:

I have a number of ceramic grade "1" block magnets from Harbor Freight, 3/8" X 7/8" x 1-7/8". I calculated the Gauss on the online "Centerline Calculator for a Rectangle Ceramic Magnet" and determined that the Gauss of the ceramic block at the centerline, at a distance of 2" was 97.56.

I attached the ceramic block magnet 2" behind the electro-magnet and it stuck to the refrigerator. I then applied 1500 mA of D.C. current to the electro-magnet and it neutralized the attraction strength completely, but not to the inside.

This test confirms the Gauss calculation I performed for the air core bifilar as perfectly correct at 113. Gauss is determined very simply by multiplying Amps times Henrys of Inductance; Now that's final; Case closed!
__________________
 

Last edited by Allen Burgess; 10-20-2017 at 04:25 PM.
Reply With Quote
  #178  
Old Yesterday, 11:09 AM
Allen Burgess Allen Burgess is offline
Gold Member
 
Join Date: Sep 2011
Posts: 1,381
Simplicity of neutralization.

Amperage can be measured by placing the DMM meter electrodes in series between the positive of the power source and the positive lead of the inductor. Make sure the input does not exceed the capacity of the meter fuse. My VICI has a 20 Amp fuse. The other factor is the inductance of the inductor in Henrys.

These two factors multiplied together will produce the magnetic force in Gauss of the electro-magnetic coil.

Next, consult an online calculator and determine the Gauss of the permanent magnet on the centerline. The calculator will give you the exact size of the spacer you need to equal the Gauss of the inductor. These simple measurements can save anyone a lot of time and pinched fingers over the trial and error approach.

Once the magnet neutralization coil is assembled, your'e next step would be to fashion an overhead oscillator portion. This is much simpler. The contact would consist of two wires to short the power source through the pulse coil.

Working with a battery, this kind of oscillator can reverse charge it's own source by "Super Position". The re-gauging of the permanent magnets generates a higher voltage then the power source, and the power flows backwards. Try it, it's a lot of fun and very easy to do do with the correct math.
__________________
 

Last edited by Allen Burgess; Yesterday at 11:13 AM.
Reply With Quote
Reply

Tags
amplifier, construction, magnetic, mechanical, torque

Thread Tools

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off



Please consider supporting Energetic Forum with a voluntary monthly subscription.

For One-Time Donations, use admin@ this domain > energeticforum.com

Choose your voluntary subscription

All times are GMT. The time now is 06:24 AM.


Powered by vBulletin® Version 3.8.8
Copyright ©2000 - 2017, vBulletin Solutions, Inc.
Search Engine Optimisation provided by DragonByte SEO v1.4.0 (Pro) - vBulletin Mods & Addons Copyright © 2017 DragonByte Technologies Ltd.
Shoutbox provided by vBShout v6.2.8 (Lite) - vBulletin Mods & Addons Copyright © 2017 DragonByte Technologies Ltd.
2007-2015 Copyright - Energetic Forum - All Rights Reserved

Bedini RPX Sideband Generator

Tesla Chargers