Basic Free Energy Device

[Matthew Jones]

Hi Matt.
Here is latest. Motor is mounted and running. At 14.4 V split core comes alive and motor starts cycling with amps dropping to 1.5 and picking up again momentarly to 5-7A. At 12.5 V unloaded it pulls 4.8A. As the cores start to drag magnets with frequency in resonance split core starts to delay and recoils accelerating rotor. On the end I pull pin and separate magnetic assembly and you can see motor is pulling 1.A at 12.6V.
As for speed I think video says it all. If I push motor harder I am afraid magnets are going to separate. Your litle motor is a powerhouse. Soon I will wind ce one one and report all the measurements and wier sizes so to clear any misunderstanding or miscalculation if it is there.

https://youtu.be/QGPwy6eF_UU

Like I said no ones doubting its abilty I would just like confirmation on the RPM's. If you could tach that gen that would be good enough. Previously you were using a sticker wrapped around the shaft and seems to throw off my tach. So if you can in anyway use a spinning object and get an rpm that would be great. We just want to confirm.
This way if the motor is capable of 5300 rpms as you showed before then I can set the driver to that speed but as of now I have not been able to hit those speeds at that voltage with that current draw. Its all about trying to make this driver as accommodating and flexible as possible.

Thanks
Matt

[desa]

Hi Matt.
You are absolutly correct. It is error in measurement. Sticky tape does trove this meter off. I remeasured using tiny sticker for laser reflectors and instead of original 3600 I am getting 2100 rpm at 12.8 V at 1.0A what is in line with your findings (motor running alone). Sorry about that. To get to 3100 rpm I must go to 18 V end 1.2A. Unloaded gen attached is raining at 1620 rpm with 4.8A at 12.9 V. Thank you for pointing it out it definitly makes difference.
David.

Like I said no ones doubting its abilty I would just like confirmation on the RPM's. If you could tach that gen that would be good enough. Previously you were using a sticker wrapped around the shaft and seems to throw off my tach. So if you can in anyway use a spinning object and get an rpm that would be great. We just want to confirm.
This way if the motor is capable of 5300 rpms as you showed before then I can set the driver to that speed but as of now I have not been able to hit those speeds at that voltage with that current draw. Its all about trying to make this driver as accommodating and flexible as possible.

Thanks
Matt

[voltan]

R.I.P. J.B and go you good thing Donald Trump.

[Matthew Jones]

Hi Matt.
You are absolutly correct. It is error in measurement. Sticky tape does trove this meter off. I remeasured using tiny sticker for laser reflectors and instead of original 3600 I am getting 2100 rpm at 12.8 V at 1.0A what is in line with your findings (motor running alone). Sorry about that. To get to 3100 rpm I must go to 18 V end 1.2A. Unloaded gen attached is raining at 1620 rpm with 4.8A at 12.9 V. Thank you for pointing it out it definitly makes difference.
David.

Thanks for checking that..Mistakes happen its not a worry.

Currently under the winding pattern that I outlined originally we are getting about 3200 rpms at 12-13 volt 1.2 amp. 25 volt at 1.9 amp delivers 4400 to 4500 rpms if the motor is equipped with good bearings (After market).

So in light of what I said before if anyone is building this motor stick to the original plan. I was ready to endorse a new winding after confirming but this did not work out. I did do alot of work to optimize the pattern. One thing may be incorporated is litzing the wire.

Currently we are using about 60-70 ft (20 -22 metters) 3 strands of #24 awg (.51 +- mm) wire. Again litzing the wire may turn out better results.

Driver and gen coming real soon.

Matt

[Turion]

Wire Length

Matt,

Just so you know. I used three strands of #24 that were 45 feet long to do 40/30/30, and had a few feet left over. This was on the SMALLEST razor scooter motor. The larger ones will use more wire.

Dave

[Matthew Jones]

Ya the smaller motors are going to take less. The real point is to stay under 1 ohm of resistance in your wire.
So with 65ft of 24 awg you get 1.66 ohms divided by 3 you get .55 ohms.

Technically you could push 45 amp through that so the only thing that is limiting current is the on off time of the motor. When the coils completely discharge every 90 degrees the rotor is always in a surge state or as much as it can be.

We can control the amount of current getting to the coils externally and collect and reuse what comes back out with just little cost to the torque.

Matt

[jettis]

Now perhaps some one out there like you Dave, Matt or Duncan or anyone ,could you please help me to understand why adding independent resistive loade on battery 3 would increase motor speed (now running at 5300 RPM ).

Desa,To answer your question above... adding a load to battery three would increase the speed of the motor between the positives simply because the voltage of battery 3 is decreasing and therefore creating a larger voltage difference (differential or potential difference) between the two positives. More voltage differential equals more voltage seen across your motor which means more motor speed. The opposite applies too... less voltage differential between the positives, for the motor, equals lower motor speed.

If we had 24 volts (batteries 1&2) on one side and a 12 volt fully charged battery (battery 3) on the other side and connected the grounds together and put a DC motor across the positives we would see... The 12 volt fully charged battery increase in voltage from its resting voltage to 15-16 volts fairly fast and the motor would slow down as the voltage difference between the 24 volt battery and the 12 volt battery becomes increasingly less. We would see 24 volts - 16volts = 8 volts across the DC motor that is running between the positives.

Now let's say you put a big enough load across battery3, while it's charging from the DC motor between the two positives that it will drop the voltage on the third battery from 16 volts down to 13 volts. Which leaves us with 24 volts - 13 volts = 11 volts across our DC motor between the positives. Hence the DC motor will run at an increased speed (when compared to the 7 volts across the DC motor from the above example) on the 11 volt potential difference across the DC motor that is connected between the positives of batteries 1 & 2 and the 3rd battery.

I hope that answers your question.

Dave Wing

[desa]

Hi Jettis.
Yes I see it now. Thank you for taking time to answer. I took the measurements and it corresponds perfectly. It answers why battery 2 drops more than batery 1. At the moment in testing setup my battery 3 goes to 16 volts and I have to stop to let system equalizes before continuing again. I am afraid in damaging my new batteries and specialty because thay are gell mat. This is reason I was adding resistive load. I dislike swapping arrangement so It would be nice to come up with some solution to this predicament. It is nice to be in situation we have returning power as a surplus.
David.

Desa,To answer your question above... adding a load to battery three would increase the speed of the motor between the positives simply because the voltage of battery 3 is decreasing and therefore creating a larger voltage difference (differential or potential difference) between the two positives. More voltage differential equals more voltage seen across your motor which means more motor speed. The opposite applies too... less voltage differential between the positives, for the motor, equals lower motor speed.

If we had 24 volts (batteries 1&2) on one side and a 12 volt fully charged battery (battery 3) on the other side and connected the grounds together and put a DC motor across the positives we would see... The 12 volt fully charged battery increase in voltage from its resting voltage to 15-16 volts fairly fast and the motor would slow down as the voltage difference between the 24 volt battery and the 12 volt battery becomes increasingly less. We would see 24 volts - 16volts = 8 volts across the DC motor that is running between the positives.

Now let's say you put a big enough load across battery3, while it's charging from the DC motor between the two positives that it will drop the voltage on the third battery from 16 volts down to 13 volts. Which leaves us with 24 volts - 13 volts = 11 volts across our DC motor between the positives. Hence the DC motor will run at an increased speed (when compared to the 7 volts across the DC motor from the above example) on the 11 volt potential difference across the DC motor that is connected between the positives of batteries 1 & 2 and the 3rd battery.

I hope that answers your question.

Dave Wing

[Matthew Jones]

By adding a load across battery 3 you are reducing the resistance. @15v your batteries internal resistance is pretty high because your talking about the resistance between the positive pole and the ground pole. And your potential difference is really small so you do not have the ability to push much current.

So then you parallel another load to battery 3 like the resistor which is dumping power to the ground pole and that lowers the overall resistance of the system. So you no longer are solely dependent on the batteries ability to sink or transform the current.

There is nothing wrong with what you are doing but I am not fan of it. Its just burning power. Fortunately in your case and others enough power is still making it through the system on the ground side to keep the primaries going while you hold up battery 3. It would be better to catch all that power and distribute it back to the system. With Battery 3 at 15 volt and an 11.2 ohm resistance your burning off just over 20 watts. Thats kinda big considering your motor motor cost nothing. Whats saving you is that motor oscillating faster. The faster oscillation are now going negative at a higher rate and the primaries keep there potential.

I explained how power travels in the system before but I'll go through again.

You can look at a battery/bank 2 ways if you measure it with the red lead on the red pole you will get positive. If you flip the lead you read a negative voltage. So depending on what point of time you are at when running one of these systems and depending how charged the third battery is. Every time your motor pulses, the pulse on the wire PLUS the charge in the third battery is higher in potential than the primaries. So you get a pulse that goes from the negative of the third battery up to the negative of the primaries that is at a higher potential than the primaries and your primaries appear not to discharge. But in fact all your doing is circulating energy in a DC loop while maintaining the separate dipoles.

Its kinda hard to grasp but once you see it happening and understand how everything is flowing then you can start to grasp whats possible.

I think instead of burning that extra power I would move it back upstairs. You get the benefits of the potential system and you get more control over the potential. Ideally you might be able to charge all the batteries up to 14-15 volt and everything in the system would become more efficient while delivering the ability to do more work.

Matt

Hmmmm??? Thought I laid this out, must have made it too confusing. I'll work on that.

LOL
Matt

[jettis]

Hey Matt,

When a fully charged lead acid battery, that is in prime shape, it has a very low internal resistance. Now when you load the battery and take some capacity out you are increasing the internal resistance within the battery, hence the voltage drop.

So with that said I do not follow your line of thinking... Or am I missing what you are saying?

Dave Wing

Hmmmm??? Thought I laid this out, must have made it too confusing. I'll work on that.

LOL
Matt

[Matthew Jones]

So your saying it easier to charge a full battery than it is to charge an empty battery?
Remember I am talking about charging the battery. When your discharging a battery the voltage doesn't drop because of resistance it drops because there is less charge on the positive plates. Internal resistance comes into play only as the capacity of the plate is reduced the battery cannot take or deliver as much power. Lower capacity higher resistance to taking or delivering a charge depending on what causes the lower capacity.

A full battery has to dissipate any incoming charge into heat and that is the highest form of resistance.

Matt

[jettis]

In my opinion it is easier to charge a full battery, but it may potentially waste more energy because their comes a point when the battery cannot receive any more charge and you begin to start wasting energy.

Now in reference to your reply below here is a link... I believe Peter posted or at least referenced this some time ago. Read chapter 5 for starters, it talks about what we are discussing currently.

http://www.pureenergysystems.com/PES...le.pdf#page443


Dave Wing

So your saying it easier to charge a full battery than it is to charge an empty battery?
Remember I am talking about charging the battery. When your discharging a battery the voltage doesn't drop because of resistance it drops because there is less charge on the positive plates. Internal resistance comes into play only as the capacity of the plate is reduced the battery cannot take or deliver as much power. Lower capacity higher resistance to taking or delivering a charge depending on what causes the lower capacity.

A full battery has to dissipate any incoming charge into heat and that is the highest form of resistance.

Matt

[Matthew Jones]

Well I guess I can't argue with Bedini, but since I purchased a battery analyzer several years I have seen a different picture.

A healthy battery has a very dynamic internal resistance.

Matt

[jettis]

Well I guess I can't argue with Bedini, but since I purchased a battery analyzer several years I have seen a different picture.

A healthy battery has a very dynamic internal resistance.

Matt

Well are you going to explain? I'm interested.

Dave Wing

[Matthew Jones]

Look just take any battery you careless about outside, dump a regulated 20 amp on it while its discharged, watch the meter, as the voltage peaks it will no longer take 20 amp. That happens because as the battery fills up it has to dissipate the current into heat. The battery is no longer capable of storing power.
Now you wanna see that in the opposite direction, put your meter across a the poles of the battery with a dead short or as low of resistance as you can. As the potential on the plates drops so will the current. This is based on ohms law but as the potential drops the resistance internal to the battery goes up as well which you see by calculating ohms law. But if you reverse the current at this point you reverse the resistance until its full again + Damage.

See they work well at a medium range but at either end of the spectrum the resistance is high, its dynamic. Lots of variables.

Thats why with the 3 battery setup when a battery is charged high the current slows down. You can calculate ohms law and see some the current is slowed from the low potential difference and then you can also see the internal resistance to the battery factors above that.
So for instance you have 10 volt of difference and 1 ohm load you should pass 10 amp. But if that 3rd battery is full you may only be able to pass 2 amp. Where is the resistance coming from?
At that point you have another 4 ohms of resistance. A total of 5 in the system and you ad 10 ohm resistor across batt 3. Now your down to 3.3 ohms, more current can flow and good portion of it now is being turned to heat outside the battery, which in itself will lower the internal resistance inside the battery.

Its Dynamic. Static internal resistance has to do with the amount of power a battery will DISCHARGE by design. But there are Dynamic properties at play during either the charge or discharge cycle and those have to accounted for while in use.
Even though battery is a negative resistor and can technically take as much power as you give it until it destroys itself resistance is the factor in that. Heat, and how much of it can we dissipate.

Matt