So in light of UFOs being demonstrated well enough to exist for me to accept they are real and they fly, I've become interested in figuring out how to get the energy needed to do what they do.
Doing this has led me down the rabbit hole of reading old patents and old theory to see if there is a branch of science that got neglected for some reason or another.
In the investigation, it leads me to electrostatic machines.
Now these are interesting.
They are typically made using 1 nF Leyden jars and they can make megavolts of electricity with the right set up.
Regardless, I'm watching this YouTuber turn his machine by hand and it starts throwing 6 inch sparks every 2 seconds.
I ran the numbers and that's 457 kilovolts to jump that far. At 1 nF that works out to 200 watt-seconds per spark.
So an average of 100 watts per second as an average.
He's spinning it with his hand, when he stops, it takes a good 15 seconds for friction to stop the wheels. I have serious doubts that something that could be spin with one extended finger uses 100 watts to run.
The motion of the wheels are really only being affected by friction. In span the same way in an early part of the assembly video when none of the electrical components were attached.
This is an electron pump in the same way as an air conditioner has a heat pump.
I think high voltage electrostatics is the key to efficiently separating charge so that you can do other work. There's basically almost nothing for me to research about these things. They aren't even conspiracy level, they're just assumed useless outside of being a demo of high voltage and abandonned as a technology.
The only place I can see where they use it is in high voltage experiments, everything always comes with the discalmer that the watts made by these machines are uselss because it's high voltage and low amperage, like step down transformers and power conditioners don't exist.
They are also super, rediculously dangerous with big capacitors and can kill you in an instant by striking you with lighting from a distance of inches!
Edit:
- Edited because I wrote uF instead of nF.
- Added link to video series on how to make this particlular type of electrostatic machine from beggining to end including a working demo. (I assume he makes them as demo units for universities) https://www.youtube.com/watch?v=3kMQJk8HZZg&t=2899s
- A comparison of how hard it is to make 700 watts of power through electromagnetics https://www.youtube.com/watch?v=S4O5voOCqAQ
- Calculations by me can be found in the comments bellow
Watts per second???
I'm not sure what you are calculating here. Watts are a unit of power, not energy. Are you trying to say joules per second? (i.e. 100 watts) Or are you trying to say watt-seconds, which is a unit of energy? (albeit a tiny one...1 3.6 millionth of a kWh, the units of your electrical bill)
If you could post your mathematical calculations it might make more sense, and allow us to offer feedback. I am having trouble following what you are trying to say with only the above information.
1 Farad = 1 amp seconds worth of power supplied at 1 volt
an amp-second is known as a coulomb
1 Farad = 1 coulomb stored at 1 volt
since 1 coulomb = 1 amp-second, a farad is 1 watt-second
Wattage is power, power and energy are synonyms in my vernacular; they aren't called energy plants, they're called power plants. Farad a description of ability to store power. Watts are typically given in watt-hours.
The capacitance of a capacitor is the amount of charge that can be stored per volt of applied voltage. So, if a capacitor is rated at 1 nanofarad (1 nF), then the amount of charge it can store at a given voltage can be calculated as follows:
Q = C * V
where Q is the charge stored on the capacitor in coulombs, C is the capacitance of the capacitor in farads, and V is the voltage across the capacitor in volts.
So, if a 1 nF capacitor is charged to 450 kilovolts, which is equal to 450,000 volts, the amount of charge stored on the capacitor can be calculated as:
Q = C * V = 1 * 10^-9 * 450,000 = 0.00045
So, the 1 nF capacitor will hold 0.00045 coulombs of charge at 450 kilovolts.
So that's 0.00045 amp seconds being pushed by 450 kilovolts.
0.00045 A/S * 450 KV = 202.500 KV-A/S (or watt seconds)
That's assuming a 1 nanofarad jar. Consider that a typical Leyden jar of one pint size has a capacitance of about 1 nF.
That would also explain why the sparks looked ~5 mm wide in the video.
https://www.youtube.com/watch?v=3kMQJk8HZZg&t=2899s
I think it's like a heat pump for electrons.
It's like how we can use a resistor wire to create heat and make 1 watt of heat for every 1 watt of electricity put in the wire. But if you use state changed instead, you can relocate and concentrate 3 watts of heat by spinning a compressor with 1 watt of power.
A typical generator uses the magnetic force to move electrons around. This is susceptible to back-emf and that back emf pushes back against you.
With this, the EMF you generate is really strong for the ammount of work put in, so with the right conditions, that EMF can push a lot of current and you don't have to fight against the back EMF.
Watch the video. The thing makes 450 KV sparks while slowing down to zero over 10-15 seconds.
It's not hard to turn this thing.
I'm in the process of making one right now actually, I want to see this for myself in my own home.
Look at how hard this olympic level athlete has to work in order to push 700 watts with electromagnetics. https://www.youtube.com/watch?v=S4O5voOCqAQ
compare that to the man in the video turning the wheel with one hand all relaxed.
Let me see if I can help. You have 1nF cap with a working voltage up to 450kV. (You will see later the capacitance of the jar is actually irrelevant to the energy used.) But that voltage will never be reached in the current setup. Once the voltage climbs up above the breakdown resistance of the air between the balls (take an upper bound of 10kV/cm) a spark will form. The air basically forms a capacitor between the 2 balls. At 6", the breakdown voltage of this air capacitor is around 150kV. So that is the highest voltage you can ever reach. It's like having 2 capacitors in parallel. Once the breakdown voltage of the air capacitor is reached, the jar capacitor can't climb any higher because current just discharges through the air.
At this point, there will be 1/2 * C * V^2 in the capacitor (or 0.5 * 10^-9 * 150,000^2 = 11.25 joules of energy in the capacitor.
(Side note: in your calculation, you used QV, which is wrong. The correct formula is 1/2QV because you need to use the average voltage during charge. The capacitor starts at 0 volts and charges up to maximum. So the actual stored energy is only half the final voltage. The formula above 1/2CV^2, is the same formula you used, just substituting Q=CV)
So far, I get what you are saying, and it is basically correct up the factor of 2 and the breakdown voltage of the E field in air.
Here is where I can't follow your logic. This is what happens after 150kV is reached:
We push the voltage slightly higher, and a spark forms. The air is ionized between the 2 balls, and a current flows. This will continue until the E field in the air gap dielectric drops below the point it can keep the air ionized. At that point the current will cease and the spark will end. Power is only lost during the brief time the spark is active, so that is the only power that needs to be replaced by the operator spinning the wheel.
In order to proceed further and determine what that power is, we need a model of the air gap capacitor. The actual energy dissipated is defined as the integral of VI dT. Unfortunately both V and I are time varying functions, so we need a model. I couldn't find one, but I did find some rules of thumb on a spark plug forum that the maximum energy it is possible to dissipate in a spark being about half of the energy stored in the gap capacitance. So let's use that. If we assume the spark is 5mm wide, then we can approximate the capacitor as a 5x5 mm^2 plate capacitor separated by 6"(152mm). Air has an epsilon = 8.8510^-12, and the total capacitance is 8.85*10^-12 * (0.005)^2 / .152 = 1.45 * 10^-15 or 1.45 fF (femtofarads). (Using the formula for parallel plate capacitor C = epsilon * A / d)
This is the actual capacitor controlling the energy discharge of the circuit on each spark. So how much energy is stored in a 1.45 fF capacitor at 150kV? Again, that is 1/2CV^2, or around 16 micro joules. Half of that would be around 8 uJ (microjoules).
So each spark only uses around 8 uJ of energy. What does this mean:
Before the spark the jar contained 11.25 joules of energy. After the spark, the jar contained 11.249992 joules of energy. Almost nothing was lost to the spark. So the jar only needs a very small additional input of energy to create additional sparks. Even at 10 sparks per second, that is only around 80 micro watts. That is an incredibly tiny amount of energy, easily created by a finger.
So I don't think there is anything unusual happening here. But it is a cool project and presentation.
You're right, I forgot to average the voltage. However you forgot that an uncharged capacitor present iteslf to the circuit like a open short.
You have to charge the capacitor to the breakdown voltage in the air like in a marx generator (https://www.youtube.com/watch?v=dje7uhyW23o) which is not 10 kv/cm, it's 30 kv/cm https://www.microwaves101.com/encyclopedias/atmospheric-breakdown
So you canot escape that he his charging a >0.5 nF capacitor setup to 450 KV, which takes 100 watts of energy to do.
Plus the capacitors fully discharge because if you use a bonetti style instead of having metal sectors, the capactors overshoot when discharging and flip the polarity of the entire machine and makes AC instead of DC.
The following example is less voltage AND less current AND it's running back through a corona motor into a dc motor and it's still powering 18 watts of light bulbs.... mind you it's coming from the sky, but it's still amps and volts and it's still less than what these machines generate by 2 orders of magnitude.
https://www.youtube.com/watch?v=vDRCKVUO8vw
I can grant you all of that, but even tripling the breakdown voltage to 30kV/cm, and allowing double the energy flow out of the air gap capacitor, it still only increases total power used by 2 * 3^2. Which increases the power per spark from 8uJ to 144 uJ. It is still such a tiny amount that it really makes no difference. The whole system with sparks at 10 Hz is still on the order of a milliwatt.
Sadly it's just very hard to be accurate without a realistic model of the E field between those 2 balls. That is what you need to spend your time investigating, because that is your load in this context. The E field completely controls the power utilized. Without an accurate model of that, all calculations are just guesswork. The jars and other features of the circuit are just a fancy way of making a battery.
But good luck on your project. One way or the other I'm sure it will be educational.