PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-11-26, 21:38:21
News: Registration with the OUR forum is by admin approval.

Pages: 1 2 3 4 [5] 6 7 8 9 10 11 12 13
Author Topic: Using the Earth's vector magnetic potential  (Read 102639 times)
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Smudge (and all)

I posted this a while back and wonder how it might fit the current ideas of electrostatic devices on a spinning disc.

http://www.overunityresearch.com/index.php?topic=415.msg6604#msg6604

Spinning coils would develop a potential which is occasionally shorted when the  tube ionizes.

Some musing on this would be interesting.
« Last Edit: 2017-02-08, 02:01:01 by ION »


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

Group: Moderator
Hero Member
*****

Posts: 1940
Hi Ion,

If you consider the coils as bodies that can have excess charge then it is possible they could receive force impulses that are torque impulses from the Earth's A field if the charge is switched in synch with the rotation.  The synch could be obtained from the coils intersecting the Earth's horizontal B field.  The high voltage could come from the flourescent starter.  But it would need more than just the starter, e.g. an energy source like a storage capacitor and some reasonably complex switchery that would not be available at that pre-semiconductor time.  Maybe the switching could be done by spark gap with the induced voltage from the coils which would be maximum when they are aligned E-W, but that would require the axis of the coils to be at right angles to the diameter joining them (that diameter must be N-S at the switching point). 

Smudge
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
Dear Cyril.

In engineering there are many ways to " skin " the proverbial! I have worked out a plan to create a large diameter plastic disc without a lathe.

I'm intending to use those 2" Brass balls, of which I have many. My intention is to " press fit " them around the edge, my PCD being an 1,1/4" less than the OD.

What would be the ideal separation distance between one ball and the next?

Kind regards, Graham.


---------------------------
Nanny state ? Left at the gate !! :)
   

Group: Moderator
Hero Member
*****

Posts: 1940
Dear Graham,

Not sure what is the best separation, but if you put in 16 spheres you would get just under 3/4 inch if my math is correct.  I would think that is about right.  However having just read Jefimenko's paper at http://rexresearch.com/jefimenko/jefimenko.htm you might consider just using the plastic disc and placing charge on its surface using corona discharge from a comb that has sharp teeth.  If you look at Poggendorff's motor Figure 28 he has his combs at an angle so as to make it self starting and unidirectional, see first paragraph on page 59.  If you had combs along the radii with their teeth normal to the surface the motor can run in either direction after being given a push.  Then you look for different speeds in the two different directions.  What do you think?  My worry with the spheres is the total weight that puts load on the bearings and it might not run at all, but I am not really skilled in mechanical engineering and might be a bit paranoid there.

Best regards

Cyril
   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Dear Cyril.

In engineering there are many ways to " skin " the proverbial! I have worked out a plan to create a large diameter plastic disc without a lathe.

I'm intending to use those 2" Brass balls, of which I have many. My intention is to " press fit " them around the edge, my PCD being an 1,1/4" less than the OD.

What would be the ideal separation distance between one ball and the next?

Kind regards, Graham.

Dear Graham,

For cutting large discs I use a radius arm attached to my router and use a fine point vee bit, then cut both sides about halfway through the material. Sometimes I experiment with the speed of the router so I don't get melting. I'm sure you have done this in the past, I'm just thinking out loud.

Also cutting a small circular groove to accept the edge of the hemi-spheres where the they are to be bolted will keep them from walking under high speed. If this is done on each side of the disc it should be very rigid.

I'm a little fuzzy on what is needed here. I imagine a large sphere will have more capacitance and more of a surface effect that cannot be obtained by just using doorknob capacitors inside smaller spheres for more capacitance. Or is this thinking wrong?

Kind regards
ION


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

Group: Moderator
Hero Member
*****

Posts: 1940
The self capacitance of a sphere is proportional to its radius r and is given by C=4*pi*epsilon*r where epsilon is the permittivity of free space 8.85pF/m.  For simplicity the capacity is close to 1 pF for every 1cm of radius.  So Grum's 2 inch spheres will have a self capacitance of 2.5pF.  The charge it gets will be related to the voltage by Q=CV.  The change of charge that occurs at each brush is then 2CV (and that's not the French car :)).   The force impulse (force*time) is 2*C*V*A where A is the vector magnetic field.  Taking 100 Weber/m as the Earth's vector field Grum's wheel will get a 2*2.5*10-10*V force impulse at each passing.  With his radius being 17.5 cm his wheel will get get a torque impulse of 1.75*V Newton-meters-seconds.  With 16 spheres and rotating at R revs per second that is an average torque of 2.8*10-9*V*R Newton-meters.

Smudge
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
Dear Cyril.

Ok.... 14" it is...   O0

I was intending to use my Rotary table under the milling machine.

I might also try the Poggendorff's motor approach first !!   ;)

Dear ION.

It might surprise you to learn that a Router is a tool I don't have. Probably because my milling machine can perform the same function, but not so easy to carry around !!   :)   It's a good idea though, worthy of mention to those that don't have bigger kit.

I took myself to the workshop today to oversee my son's and grandson doing some casting..... Hmmm, they've got a lot to learn. A couple of pictures attached.

Kind regards, Graham.


---------------------------
Nanny state ? Left at the gate !! :)
   
Group: Ambassador
Hero Member
*****

Posts: 4045
Grum
Looks like a good way to learn the ABC's


   

Group: Moderator
Hero Member
*****

Posts: 1940
If this hidden momentum lark actually works, then we need a means of storing then releasing charge and its hidden momentum that operates at a much lower voltage.  Here is a suggestion that it can be stored by chemical means.  I use the lead-acid battery merely as an example since I already had the ionic flow diagram on my computer.

Smudge
   

Group: Moderator
Hero Member
*****

Posts: 1940
I have updated that last paper along with a suggested experiment.  This does not need the EHT from a Van de Graff so it might suit other experimenters.  It uses electrodes cut from those salvaged from a car battery, plus electrolyte and of course relatively low voltage pulses.  The aim is to get a reasonably high current but narrow pulse into the device and look for a torque impulse.  If a force from change of hidden momentum can be found it opens up so many possibilities.

Smudge
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
The self capacitance of a sphere is proportional to its radius r and is given by C=4*pi*epsilon*r where epsilon is the permittivity of free space 8.85pF/m.  For simplicity the capacity is close to 1 pF for every 1cm of radius.  So Grum's 2 inch spheres will have a self capacitance of 2.5pF.  The charge it gets will be related to the voltage by Q=CV.  The change of charge that occurs at each brush is then 2CV (and that's not the French car :)).   The force impulse (force*time) is 2*C*V*A where A is the vector magnetic field.  Taking 100 Weber/m as the Earth's vector field Grum's wheel will get a 2*2.5*10-10*V force impulse at each passing.  With his radius being 17.5 cm his wheel will get get a torque impulse of 1.75*V Newton-meters-seconds.  With 16 spheres and rotating at R revs per second that is an average torque of 2.8*10-9*V*R Newton-meters.

Smudge

Dear Cyril.

Could you simplify the above equation into something I can understand please? Mathematics wasn't my strongest subject.

I ventured out to the workshop but the biting NEasterly wind soon put paid to any serious work. I've ordered some 316 SS bearings and a 48 mm hole saw as the spheres aren't quite 2" in diameter.

My plan is to suspend the disc using a small DC motor as the upper bearing and use a needle point on glass as a lower preload. The motor will then be used as a Dynamo to ascertain any increase/decrease in the rotors orientation.

I will be hoping to proceed with some experiments later next week, just hoping the HVM arrives soon.

Kind regards, Graham.


---------------------------
Nanny state ? Left at the gate !! :)
   

Group: Moderator
Hero Member
*****

Posts: 1940
Dear Grum,

I think the best way to approach this is to use the current being supplied by the VDG rather than the voltage when your beast is rotating.  You can measure the current with a micro-ammeter in series with the earthy end of the connection.  Taking the earth's E-W magnetic vector potential in Wales to be about 100 Weber/m (its 200 at the equator and zero at the poles) then your disc could get an anomalous torque of value current(uA)*disc diameter(m)/10,000 Newton-metres.  1Nm=0.7376 ft-lbs.  So if you get 1uA and with your diameter (centre to centre) of 0.35m you should get 2.582*10-5 ft-lbs.  That's a weeny torque to add or subtract from the system and I have no idea as to whether it will show up as a change of speed.

Smudge
   

Group: Moderator
Hero Member
*****

Posts: 1940
I have found a patent WO2012053921 on electromagnetic propulsion that uses the change of charge in a magnetic vector potential to get thrust.  So I am not alone.  See attached extract.  I can't find any meaningful images, those available are too poor a quality.  It's looking good for a positive result.

Smudge
   

Group: Experimentalist
Hero Member
*****

Posts: 568
I have found a patent WO2012053921 on electromagnetic propulsion that uses the change of charge in a magnetic vector potential to get thrust.  So I am not alone.  See attached extract.  I can't find any meaningful images, those available are too poor a quality.  It's looking good for a positive result.

Smudge
Is he creating his own A field with the center magnet?


---------------------------
"Whatever our resources of primary energy may be in the future, we must, to be rational, obtain it without consumption of any material"  Nicola Tesla

"When bad men combine, the good must associate; else they will fall one by one, an unpitied sacrifice in a contemptible struggle."  Edmund Burke
   

Group: Moderator
Hero Member
*****

Posts: 1940
Is he creating his own A field with the center magnet?
Yes he is so I think his propulsion system won't work because he hasn't taken into account the reaction on that magnet.  But his patent is the only written work I can find that states that A*dq/dt represents a force that is real and can be used.

Smudge
   

Group: Professor
Hero Member
*****

Posts: 3499
his patent is the only written work I can find that states that A*dq/dt represents a force that is real and can be used.
Most arrangements that are responsible for the dq/dt cause a flow of charge from one place to another (electric current) and that creates its own set of force reactions.
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
Hello all.

Ok, all ordered materials are in with the exception of the Chinese HVM.

Now, can someone do a little mathematical work please?

My disc will have an OD of 355 mm. I'm going to pitch the spheres on a 335 mm PCD. This action will present a nearly hemisphere to the edge but a small portion of the 48 mm dia hole will be returning to secure the spheres.

Cyril suggested a 16 hole division. So I need to know the linear value in mm to set my dividers to. I could use my rotary table, but the blessed things too heavy for me to lift these days!

Thanks, in advance, Graham.


---------------------------
Nanny state ? Left at the gate !! :)
   

Group: Moderator
Hero Member
*****

Posts: 1940
By my calculations you set your divider to 65.355mm.  If the angular separation between centers is theta the formula is PCD*sin(theta/2).  So that is 335*sin(11.25) in this case where the centers are 22.5 degrees apart.  I hope that's right.

Smudge
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
By my calculations you set your divider to 65.355mm.  If the angular separation between centers is theta the formula is PCD*sin(theta/2).  So that is 335*sin(11.25) in this case where the centers are 22.5 degrees apart.  I hope that's right.

Smudge

Dear Cyril.

I committed a " Schoolboy error ", cap in hand !!  :-[

My PCD should have been 315 mm..... wanted 20 mm smaller ( ALLROUND  ) Tut, tut !! Must be getting old.  :)

Kind regards, Graham.


---------------------------
Nanny state ? Left at the gate !! :)
   

Group: Moderator
Hero Member
*****

Posts: 1940
Dear Graham,

Then I get 61.453mm to set your dividers to.

Cyril
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
Dear Graham,

Then I get 61.453mm to set your dividers to.

Cyril

Dear Cyril.

There's an error somewhere, the last division comes long.

We can't divide the circumference into 16 as the straight line is shorter than the curved one. I've lost my Zeus book in the workshop the formula is in there.

Kind regards, Graham.

Ok...... It's me again..... Must have set the dividers wrong!!   :-[


---------------------------
Nanny state ? Left at the gate !! :)
   

Group: Moderator
Hero Member
*****

Posts: 1940
We all make mistakes Grum.

The formula is in my previous post as PCD*sin(theta/2) where theta is the angular separation.  16 holes are set 22.5 degrees apart so you need PCD*sin(11.25).

Smudge
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
Dear Cyril and all.

I'm getting there !!   O0

Discovered this afternoon that the spheres are not all the same size !!  >:(

Time permitting I will be making a slightly tapered mandrel with some fine grit Emery cloth to open some of the holes.

The bearing is a precision SS ball race so shouldn't get sticky with any stray magnetic fields.

Cheers Graham. 


---------------------------
Nanny state ? Left at the gate !! :)
   

Group: Moderator
Hero Member
*****

Posts: 1940
Looking good Grum.  Can't wait to see some results.

Smudge
   

Group: Renaissance Man
Hero Member
*****

Posts: 2765


Buy me a cigar
Looking good Grum.  Can't wait to see some results.

Smudge

Dear Cyril.

Well...... After a couple of hours spent " honing " the holes all 16 spheres fitted beautifully.

Disappointment rules the day however. I think there's too much weight to the assembly, at least in an horizontal plane. It just won't budge.

Could be there's not enough oomph from the V De G. Do I need a direct connection to the spheres?

Kind regards, Graham. 


---------------------------
Nanny state ? Left at the gate !! :)
   
Pages: 1 2 3 4 [5] 6 7 8 9 10 11 12 13
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-11-26, 21:38:21