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EEVblog Free Energy B.S.

Started by Dave Wing, January 30, 2015, 04:16:18 PM

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0 Members and 2 Guests are viewing this topic.

Pirate88179

TK:

Very well done.  That was a very honest explanation and mini tutorial on the Bedini motor devices.

Bill
See the Joule thief Circuit Diagrams, etc. topic here:
http://www.overunity.com/index.php?topic=6942.0;topicseen

Dave Wing

QuoteAgain I thank you for providing the link, because the video I linked above is +highly significant+ and shows (in the "reveal") the method by which many other fakers have tried to bamboozle people into believing their impossible gadgets actually work as demonstrated.

Voltage is always relative to some reference level. In the diagram below, even though both battery terminals are called "positive", one is at a much lower "potential", i.e. voltage, than the other. So Batt3's "positive" is _relatively_ negative wrt the + terminal of Batt1, and hence current will flow normally through the load, powering it and charging the #3 battery. Beware, because it will likely _overcharge_ it with bad effects.

Hi TK,

I am not sure why you posted that trick video? Was it in reference somehow to what I posted in my videos?

The batteries 1&2 will overcharge #3, as you say only if the load is passing to much current through it. But if you select the load accordingly that will not happen. Just using the batteries in the configuration presented as shown, through resistive load tests, that approximately 50% can be recovered for reuse. There may be ways to improve upon this result but that will take time to determine. John Bedini also made the diagram you posted, here, however it is one I edited. See the diagram below for the apparent proper way to use this system.

Now onto the Bedini SG machine, it is not a free energy device, but is only a machine to demonstrate a concept. At best the machine is perhaps 20% mechanically efficient and when it comes to electrical effiency I have not been able to get over 50% total energy recovery. Mind you that is using and abusing my test batteries and running them at C-1 rates or more, they are damaged batteries but still hold a know capacity, as they are frequently load tested by using a CBA. I have no doubt that if I use good deep cycle batteries within their C-20 rate I would see far better than the 50% electrical recovery I see now.

What can be done to increase the mechanical efficiency of a SG? One can simply widen the core area of the coil to cover the full magnet width. You can also add a South Pole in between the North magnets. You can also make the rotor larger in diameter and longer with longer magnets and coils (similar to the widow motor). What would these changes do to the mechanical of the machine? Now if you add a heavy flywheel to the mix how much energy storage does one have now, that would be available for use?

-Dave Wing

MileHigh

Dave:

I don't know how you define the mechanical efficiency.   For me when the rotor is spinning the all the mechanical power is used to make the rotor spin.  There is no useful mechanical output.

MileHigh

John.K1

Quote from: MileHigh on February 12, 2015, 03:35:50 PM
Dave:

I don't know how you define the mechanical efficiency.   For me when the rotor is spinning the all the mechanical power is used to make the rotor spin.  There is no useful mechanical output.

MileHigh

I would say you can spin motor fast and be able to stop it by hand, or you can spin the motor same fast and it will pluck your hand if you try to stop it :)  I would say some ratio of the input energy and the torque is what you looking for? 

TinselKoala

If the rotor is not driving a load, MH is right: all the power driving the rotor is used offsetting the losses in bearings, windage, etc. How do you determine the mechanical power dissipation of the rotor? Like this:

You can calculate the rotational Moment of Inertia of your rotor using careful measurements of shapes and masses, and geometry.  Using the MoI you can know just how much energy in Joules is stored in the rotation at a given RPM or angular velocity. Then you can do timed, unpowered rundowns, plotting angular velocity or RPM against time. Obviously you dissipate the entire stored Joules in the entire time of the rundown, so that gives you an "average" power dissipation over the time period.  The instantaneous slope of this rundown plot at any given RPM will then give you the power dissipation in Watts (milliWatts more probably) at that RPM.  So now you can calculate the "mechanical efficiency" of the spinning rotor. If it is spinning at a steady speed, consult your rundown graph and read the power dissipation from the graph. This is the power it is receiving from the drive circuit in order to keep running steadily at that speed, offsetting the losses due to windage, bearing friction, generator effect, etc. (If it was getting more power it would be accelerating, if less, decelerating. Duh.)
If you have generator coils you can do the nonpowered rundowns with generator coils active, and inactive, and this will tell you the "cost" of generating, at any given RPM.  You should do 10 or 20 timed rundowns in each condition so that your data gets a good average, evening out glitches and inevitable experimental errors in uneven conditions.  This might take an entire afternoon of actual hard work, and require either a chart recorder or sustained attention to a stopwatch and a quick pencil. Especially since a heavy rotor on good bearings can take a _long_ time to run down. I've worked with some ( my Mylow testbed for example) that take an hour or more to run down, unpowered, from a quite modest initial spin. So getting enough data to get a good average can take a while.

Mechanical efficiency is then the ratio of the total input power to the power dissipation of the rotor as found using the above procedure.