The bistander thread

[Turion]

[Aaron]

proxy?

Aaron,

Here ya go.

Regards,

bi

}edit{

When I switch WiFi off and use LTE, I get about 1 or 2 minute to post or edit, then then trouble starts.

Are you using a proxy?

[Aaron]

myth of conservation of energy

I am taking issue with these statements from Turion.

bi

There no such thing as the conversion of energy from one form to another.

Reread my long post for the proper definition of energy. It is work, which comes about by the dissipation of potential by a resistance, period.

What form does the energy come in? Energy is not the heat, we can say it is energetic, which is a generic term for the movement of something, but quite literally, energy is not the heat itself, nor is it light or anything else. Energy is the activity the potential (EMF) experiences while dissipated by a resistance to a lower potential.

That potential dissipates back to equilibrium right there at the space that the resistance is at, which dissipated it.

A source potential is there, it gets dissipated and we witness work or energy being manifest (creation and destruction of energy simultaneously), and when there is a different "form" of work being done, it did not come from the original source potential. The original source potential was used to do work in one situation. Then, another dipole or potential difference is created electrically, mechanically, chemically, etc. and when that new dipole is created, brand new source potential enters to be dissipated to do other work in another situation, which is NOT the same potential that did the work in the original dipole.

Nothing was conserved as there is no such thing as the conservation of energy, you can't conserve an activity. And energy does not change from one form to another, just new dipoles are created that allow NEW source potential to come in to do other work. And energy only is and always is created and destroyed simultaneously whenever work is done.

You lift an object, force x distance tells you how much work was done to lift the object to a certain height. At the peak of the lift, ZERO % of that work done was stored in the object because its impossible to store an activity that potential experiences. And you can't store something was just "used up."

All you did is create a potential difference or a new dipole.

mgh does NOT tell you how much potential energy you stored in the object.

mgh or mass x gravity x height only tells you how much NEW, FRESH Gravitational Potential will enter the system from the dynamic, flowing downward gravitational force, when the object is allowed to fall.

Therefore, none of the energy used to lift the object was conserved or stored and the work done when the ball is dropped came from free, gravitational potential energy - NOT from the work done to lift the ball.

We expended x joule seconds of real work to lift the object, gravity came in with its own potential to cause more joule seconds to be done so total work done is twice as much as what we had to contribute.

If you burn a gallon of gas in a car to drive up a hill and turn the car around, you didn't conserve or store any of that potential from the gasoline in the car. You created a new dipole and when you let off the brake and coast down the hill, that work is done completely by newly sourced gravitational potential, which is now doing real work. No conservation of energy, no changing forms of energy from one form to another, etc.

[Aaron]

How capacitors are charged

When a capacitor is charged, it is not done by piling electrons on a plate.

A capacitor stores dielectricity, which is the same substance as the EMF or Heaviside Flow - it is the polarized aether that was made asymmetrical by the potential difference between two terminals on a dipole. The capacitor literally is a gas canister filled with the aether gas. You over charge a cap and you see how leaky it is - it is literally leaking a gas just like a leaky air tank or any other tank of gas.

Read this pdf: http://ericpdollard.com/wp-content/u...ic-dollard.pdf

Eric would use the term energy very differently these days due to his relatively recent research into Oliver Heaviside's work, but suffice to say, it paints the picture.

[BroMikey]

Yes but you are working from the premise "Already been done" people
who have not tried this circuit still copy and paste so called book smarts
as a defense mechanism for their insecurity.

What needs to be done is what teacher did in 6th grade, do a demo
which conclusively does away with the parroted rhetoric. Ya know that
junk about conserving the environment not joules.

[dragon]

Dragon,
I have no doubt that your experiment using 1 farad caps is accurate. Larger caps and the efficiency will go up.

In your example you sent 100 joules through the motor and recovered 50% of it. The actual recovery rate should be MUCH higher, but even in YOUR example you see no value in recovering HALF of the energy used by the motor to do work?

In that example we actually sent 150 joules through the motor and captured 50 joules in the recovery cap. We only recovered 1/4 of the total charge, the other 1/4 remaining in the original charged cap wasn't used.

I find the whole process fascinating and I've built a ton of projects around that theory. Highly efficient or near unity is what I tend to spend most of my time with - as well as exploiting/converting/harvesting natural energies to make up the differences.

I don't see any benefit in using larger caps - I've done similar experiments using 6 - 2600 F units in series for a total of 433 F bank. ( including your 3 battery and single battery systems ). I have several of these banks that I use for experiments. The results of the above test would be the same as the 1 F except for the amount of time it would take to discharge/charge them and the amount of joules moved. The result would still be 1/2 lost, 1/4 unused and 1/4 in the second bank.

[bistander]

Are you using a proxy?

I don't know, sorry. Pay monthly fee for internet which goes through my WiFi to my devices. Mostly use Chrome, but tried others. The site works kinda well over my phone 3, 4G or LTE. But I can't get that on my laptop, so stuck on tiny screen.

bi

[BroMikey]

So much for all the battery theater Dragon has done the deal. So it does
get more without any special circuits, interesting. The 3 capacitor systems.
Hum....
Same volts same amps going to the load regardless if is done with a
direct connection VS the splits. Well 25% recovery for openers is better
than a kick in the teeth

The question is how can we get more than 25% recovery. Or is it 25%?
We know that a fixed load will consume the same number of joules as
long as the voltage is the same during both tests.

Let's say that the second bank Dragon uses was comprised of 2 parallel
caps collecting back the same number of joules. Now we can run that
extra energy again on a smaller load using the 3 capacitor system.

Continuing that process we can take the 2 full caps run a load and collect
back 6%. Now we are up to 31% collected back from the original charge.

Is 25% the largest amount you have recovered using caps?

[Cadman]

If I may I would like to offer my view of charging & discharging DC caps.

Call it electrons, aether gas, or electric fire it’s the charge I’m talking about. We can all agree that charge put onto one cap plate causes the same amount of charge to leave the opposite plate can’t we?

If you graph it out there is a zero volt line with pos above and neg below just like an o-scope. Both plates start at the 0v line. Put 10 units of + charge onto one plate and the graph goes to +10. During that time period 10 units leave the opposite plate mirroring the + charge rate on the graph. 10 units have been put in but the potential difference between the plates is 20 units which is how the meters read it, correct? But there is really only 10 units applied and 10 available. 10 not 20 because as one unit of charge leaves the + plate, and goes to the – plate. 1 unit has traversed the circuit but the potential difference is now 18. (10 – 1) and (-10+1) so the lines of the graph are at +9 and -9. Both plates move toward the 0v line at the same time, and after 10 units have traversed the circuit the charge is 0. 10 in and 10 out.

You don’t lose half the charge because you never had it in the first place. Saying you lose half the charge is a false verbal kludge to make the formulas look right. The formulas are right as they are.
.

[Turion]

[Turion]

Response to Cadman

I don't believe you can have movement without a potential difference already existing. You create one by adding to or subtracting from a system. But it takes interaction from an outside source for that to occur.

[Cadman]

Turion,

All true. The only other way I know of is magnetically, the Hall effect.

[BroMikey]

Bro Mikey

.........but at times as much as 85%,
..... Before there was a Stealth Fighter there were the Wright brothers
at Kitty Hawk with a flight that lasted 59 seconds.

The 3 Battery system is that flight at Kitty Hawk.

You are correct, out of the 100% of the joules moved being 150 it is
1/3 that is collected in bank 2. That would be 33% savings while what the
load eats 66% or is unchanged.

[dragon]

Something to consider... did you capture the 50 joules or did it become part of the "load" circuit? Would it make a difference if you simply moved 100 joules through a load ( conventionally ) and saved the last 100 joules for later use? For instance would the motor do as much work by shifting charge to the second cap as it would by simply driving it conventionally with the same amount of energy...? Conventionally, the motor would run stronger from 20v down to 14.2 volts using 100 joules from the cap where charging the second cap with the output would give you a quickly diminishing potential as one went down while the other rises.

Which would be better if your looking for peak efficiency?

[BroMikey]

Yes that is another good question, did you form a conclusion yet?
If a cap drains out 100 joules conventionally the potential drops on
a curve and if a 3 battery system drains out 100 joules it drops on
a curve. So if the voltages are the same you should get the same
efficiency.